Inequality $sumlimits_{k=1}^n frac{1}{n+k} le frac{3}{4}$












3












$begingroup$


I recently came across the following exercise:




Prove that $$sum_{k=1}^n frac{1}{n+k} le frac{3}{4}$$
for every natural number $n ge 1$.




I immediately tried by induction, by I did not succeed as - after some trivial algebraic manipulations - I arrive at $sum_{k=0}^{n+1} frac{1}{n+k} le frac{3}{4} + text{something non-negative}$.



Thus the solution I have found goes via comparison with the integral of $1/x$: more precisely,
$$
sum_{k=1}^n frac{1}{n+k} le int_1^n frac{dx}{n+x} =ln(2n) - ln(n+1) = ln(2) + lnleft(frac{n}{n+1}right)
$$
and the conclusion follows easily as $ln 2 le 3/4$ and the last term is non-positive.



I would like to ask first if my solution is correct; secondly, do you have other solutions? I believe it should be very easy to prove.






sequences-and-series inequality fractions harmonic-numbers cauchy-schwarz-inequality







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your inequality can't possible be true as for n=1, you would have $1/2 leq 0$
    $endgroup$
    – user172377
    Aug 6 '17 at 7:57












  • $begingroup$
    Oh right. Well, let us say that we check $n=1$ by hands and that my inequality holds for $n>1$. :)
    $endgroup$
    – Romeo
    Aug 6 '17 at 8:00










  • $begingroup$
    Try factoring out an n from the denominator, then you can easily see that the limit of the sum is equal to the riemann integral $int_{0}^{1} frac{dx}{1+x} =ln 2 leq 3/4$
    $endgroup$
    – user172377
    Aug 6 '17 at 8:06












  • $begingroup$
    Of course this is true only in the limit, hence you still have work to do here.
    $endgroup$
    – user172377
    Aug 6 '17 at 8:17










  • $begingroup$
    It looks like the sum is indeed increasing with n, meaning the result is easy now. If we let $S_n$ be the given sum, then since $S_n$ is increasing, $lim_{nto infty}S_n = sup_{n} S_n = ln2$, therefore for every n, $S_n leq sup_n S_n leq ln2 leq 3/4$ as needed.
    $endgroup$
    – user172377
    Aug 6 '17 at 8:34


















3












$begingroup$


I recently came across the following exercise:




Prove that $$sum_{k=1}^n frac{1}{n+k} le frac{3}{4}$$
for every natural number $n ge 1$.




I immediately tried by induction, by I did not succeed as - after some trivial algebraic manipulations - I arrive at $sum_{k=0}^{n+1} frac{1}{n+k} le frac{3}{4} + text{something non-negative}$.



Thus the solution I have found goes via comparison with the integral of $1/x$: more precisely,
$$
sum_{k=1}^n frac{1}{n+k} le int_1^n frac{dx}{n+x} =ln(2n) - ln(n+1) = ln(2) + lnleft(frac{n}{n+1}right)
$$
and the conclusion follows easily as $ln 2 le 3/4$ and the last term is non-positive.



I would like to ask first if my solution is correct; secondly, do you have other solutions? I believe it should be very easy to prove.






sequences-and-series inequality fractions harmonic-numbers cauchy-schwarz-inequality







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your inequality can't possible be true as for n=1, you would have $1/2 leq 0$
    $endgroup$
    – user172377
    Aug 6 '17 at 7:57












  • $begingroup$
    Oh right. Well, let us say that we check $n=1$ by hands and that my inequality holds for $n>1$. :)
    $endgroup$
    – Romeo
    Aug 6 '17 at 8:00










  • $begingroup$
    Try factoring out an n from the denominator, then you can easily see that the limit of the sum is equal to the riemann integral $int_{0}^{1} frac{dx}{1+x} =ln 2 leq 3/4$
    $endgroup$
    – user172377
    Aug 6 '17 at 8:06












  • $begingroup$
    Of course this is true only in the limit, hence you still have work to do here.
    $endgroup$
    – user172377
    Aug 6 '17 at 8:17










  • $begingroup$
    It looks like the sum is indeed increasing with n, meaning the result is easy now. If we let $S_n$ be the given sum, then since $S_n$ is increasing, $lim_{nto infty}S_n = sup_{n} S_n = ln2$, therefore for every n, $S_n leq sup_n S_n leq ln2 leq 3/4$ as needed.
    $endgroup$
    – user172377
    Aug 6 '17 at 8:34
















3












3








3


1



$begingroup$


I recently came across the following exercise:




Prove that $$sum_{k=1}^n frac{1}{n+k} le frac{3}{4}$$
for every natural number $n ge 1$.




I immediately tried by induction, by I did not succeed as - after some trivial algebraic manipulations - I arrive at $sum_{k=0}^{n+1} frac{1}{n+k} le frac{3}{4} + text{something non-negative}$.



Thus the solution I have found goes via comparison with the integral of $1/x$: more precisely,
$$
sum_{k=1}^n frac{1}{n+k} le int_1^n frac{dx}{n+x} =ln(2n) - ln(n+1) = ln(2) + lnleft(frac{n}{n+1}right)
$$
and the conclusion follows easily as $ln 2 le 3/4$ and the last term is non-positive.



I would like to ask first if my solution is correct; secondly, do you have other solutions? I believe it should be very easy to prove.






sequences-and-series inequality fractions harmonic-numbers cauchy-schwarz-inequality







share|cite|improve this question











$endgroup$




I recently came across the following exercise:




Prove that $$sum_{k=1}^n frac{1}{n+k} le frac{3}{4}$$
for every natural number $n ge 1$.




I immediately tried by induction, by I did not succeed as - after some trivial algebraic manipulations - I arrive at $sum_{k=0}^{n+1} frac{1}{n+k} le frac{3}{4} + text{something non-negative}$.



Thus the solution I have found goes via comparison with the integral of $1/x$: more precisely,
$$
sum_{k=1}^n frac{1}{n+k} le int_1^n frac{dx}{n+x} =ln(2n) - ln(n+1) = ln(2) + lnleft(frac{n}{n+1}right)
$$
and the conclusion follows easily as $ln 2 le 3/4$ and the last term is non-positive.



I would like to ask first if my solution is correct; secondly, do you have other solutions? I believe it should be very easy to prove.






sequences-and-series inequality fractions harmonic-numbers cauchy-schwarz-inequality




sequences-and-series inequality fractions harmonic-numbers cauchy-schwarz-inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 31 '18 at 7:51









Martin Sleziak

45k10123277




45k10123277










asked Aug 6 '17 at 7:51









RomeoRomeo

3,07921248




3,07921248












  • $begingroup$
    Your inequality can't possible be true as for n=1, you would have $1/2 leq 0$
    $endgroup$
    – user172377
    Aug 6 '17 at 7:57












  • $begingroup$
    Oh right. Well, let us say that we check $n=1$ by hands and that my inequality holds for $n>1$. :)
    $endgroup$
    – Romeo
    Aug 6 '17 at 8:00










  • $begingroup$
    Try factoring out an n from the denominator, then you can easily see that the limit of the sum is equal to the riemann integral $int_{0}^{1} frac{dx}{1+x} =ln 2 leq 3/4$
    $endgroup$
    – user172377
    Aug 6 '17 at 8:06












  • $begingroup$
    Of course this is true only in the limit, hence you still have work to do here.
    $endgroup$
    – user172377
    Aug 6 '17 at 8:17










  • $begingroup$
    It looks like the sum is indeed increasing with n, meaning the result is easy now. If we let $S_n$ be the given sum, then since $S_n$ is increasing, $lim_{nto infty}S_n = sup_{n} S_n = ln2$, therefore for every n, $S_n leq sup_n S_n leq ln2 leq 3/4$ as needed.
    $endgroup$
    – user172377
    Aug 6 '17 at 8:34




















  • $begingroup$
    Your inequality can't possible be true as for n=1, you would have $1/2 leq 0$
    $endgroup$
    – user172377
    Aug 6 '17 at 7:57












  • $begingroup$
    Oh right. Well, let us say that we check $n=1$ by hands and that my inequality holds for $n>1$. :)
    $endgroup$
    – Romeo
    Aug 6 '17 at 8:00










  • $begingroup$
    Try factoring out an n from the denominator, then you can easily see that the limit of the sum is equal to the riemann integral $int_{0}^{1} frac{dx}{1+x} =ln 2 leq 3/4$
    $endgroup$
    – user172377
    Aug 6 '17 at 8:06












  • $begingroup$
    Of course this is true only in the limit, hence you still have work to do here.
    $endgroup$
    – user172377
    Aug 6 '17 at 8:17










  • $begingroup$
    It looks like the sum is indeed increasing with n, meaning the result is easy now. If we let $S_n$ be the given sum, then since $S_n$ is increasing, $lim_{nto infty}S_n = sup_{n} S_n = ln2$, therefore for every n, $S_n leq sup_n S_n leq ln2 leq 3/4$ as needed.
    $endgroup$
    – user172377
    Aug 6 '17 at 8:34


















$begingroup$
Your inequality can't possible be true as for n=1, you would have $1/2 leq 0$
$endgroup$
– user172377
Aug 6 '17 at 7:57






$begingroup$
Your inequality can't possible be true as for n=1, you would have $1/2 leq 0$
$endgroup$
– user172377
Aug 6 '17 at 7:57














$begingroup$
Oh right. Well, let us say that we check $n=1$ by hands and that my inequality holds for $n>1$. :)
$endgroup$
– Romeo
Aug 6 '17 at 8:00




$begingroup$
Oh right. Well, let us say that we check $n=1$ by hands and that my inequality holds for $n>1$. :)
$endgroup$
– Romeo
Aug 6 '17 at 8:00












$begingroup$
Try factoring out an n from the denominator, then you can easily see that the limit of the sum is equal to the riemann integral $int_{0}^{1} frac{dx}{1+x} =ln 2 leq 3/4$
$endgroup$
– user172377
Aug 6 '17 at 8:06






$begingroup$
Try factoring out an n from the denominator, then you can easily see that the limit of the sum is equal to the riemann integral $int_{0}^{1} frac{dx}{1+x} =ln 2 leq 3/4$
$endgroup$
– user172377
Aug 6 '17 at 8:06














$begingroup$
Of course this is true only in the limit, hence you still have work to do here.
$endgroup$
– user172377
Aug 6 '17 at 8:17




$begingroup$
Of course this is true only in the limit, hence you still have work to do here.
$endgroup$
– user172377
Aug 6 '17 at 8:17












$begingroup$
It looks like the sum is indeed increasing with n, meaning the result is easy now. If we let $S_n$ be the given sum, then since $S_n$ is increasing, $lim_{nto infty}S_n = sup_{n} S_n = ln2$, therefore for every n, $S_n leq sup_n S_n leq ln2 leq 3/4$ as needed.
$endgroup$
– user172377
Aug 6 '17 at 8:34






$begingroup$
It looks like the sum is indeed increasing with n, meaning the result is easy now. If we let $S_n$ be the given sum, then since $S_n$ is increasing, $lim_{nto infty}S_n = sup_{n} S_n = ln2$, therefore for every n, $S_n leq sup_n S_n leq ln2 leq 3/4$ as needed.
$endgroup$
– user172377
Aug 6 '17 at 8:34












3 Answers
3






active

oldest

votes


















4












$begingroup$

By C-S
$$sum_{k=1}^nfrac{1}{n+k}=1-sum_{k=1}^nleft(frac{1}{n}-frac{1}{n+k}right)=1-frac{1}{n}sum_{k=1}^nfrac{k}{n+k}=$$
$$=1-frac{1}{n}sum_{k=1}^nfrac{k^2}{nk+k^2}leq1-frac{left(sumlimits_{k=1}^nkright)^2}{nsumlimits_{k=1}^n(nk+k^2)}=1-frac{frac{n(n+1)^2}{4}}{ncdotfrac{n(n+1)}{2}+frac{n(n+1)(2n+1)}{6}}=$$
$$=frac{7n-1}{2(5n+1)}<0.7leqfrac{3}{4}.$$
Done!



I think it's interesting that $ln2=0.6931...$.



C-S forever!!!






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Interesting is that $frac{7n-1}{2(5n+1)}=frac 7 {10}-frac 6{25n}+Oleft(frac{1}{n^2}right)$ to compare with $S_n=log (2)-frac{1}{4 n}+Oleft(frac{1}{n^2}right)$ I gave in a deleted answer based on the asymptotics. Look how close to each other are the coefficients. By the way, $to +1$.
    $endgroup$
    – Claude Leibovici
    Aug 6 '17 at 14:54





















2












$begingroup$

$$sum_{k=1}^{n}frac{1}{n+k}leq sum_{k=1}^{n}frac{1}{sqrt{(n+k)(n+k-1)}}stackrel{text{CS}}{leq}sqrt{nsum_{k=1}^{n}left(frac{1}{n+k-1}-frac{1}{n+k}right)}$$
immediately leads to $H_{2n}-H_n = sum_{k=1}^{n}frac{1}{n+k}leq frac{1}{sqrt{2}}<frac{3}{4}$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    All you really need to do is change the lower limit in the integral from $1$ to $0$, which gives the simpler result



    $$sum_{k=1}^n{1over n+k}leint_0^n{dxover n+x}=ln(2n)-ln n=ln2$$



    One way to see that $1$ is the wrong lower limit to use is that, in general, the number of terms in the sum should equal the difference of the upper and lower limits in the integral. That's because the integral comparison test usually compares each term with the area beneath a curve over a unit segment.






    share|cite|improve this answer









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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      By C-S
      $$sum_{k=1}^nfrac{1}{n+k}=1-sum_{k=1}^nleft(frac{1}{n}-frac{1}{n+k}right)=1-frac{1}{n}sum_{k=1}^nfrac{k}{n+k}=$$
      $$=1-frac{1}{n}sum_{k=1}^nfrac{k^2}{nk+k^2}leq1-frac{left(sumlimits_{k=1}^nkright)^2}{nsumlimits_{k=1}^n(nk+k^2)}=1-frac{frac{n(n+1)^2}{4}}{ncdotfrac{n(n+1)}{2}+frac{n(n+1)(2n+1)}{6}}=$$
      $$=frac{7n-1}{2(5n+1)}<0.7leqfrac{3}{4}.$$
      Done!



      I think it's interesting that $ln2=0.6931...$.



      C-S forever!!!






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Interesting is that $frac{7n-1}{2(5n+1)}=frac 7 {10}-frac 6{25n}+Oleft(frac{1}{n^2}right)$ to compare with $S_n=log (2)-frac{1}{4 n}+Oleft(frac{1}{n^2}right)$ I gave in a deleted answer based on the asymptotics. Look how close to each other are the coefficients. By the way, $to +1$.
        $endgroup$
        – Claude Leibovici
        Aug 6 '17 at 14:54


















      4












      $begingroup$

      By C-S
      $$sum_{k=1}^nfrac{1}{n+k}=1-sum_{k=1}^nleft(frac{1}{n}-frac{1}{n+k}right)=1-frac{1}{n}sum_{k=1}^nfrac{k}{n+k}=$$
      $$=1-frac{1}{n}sum_{k=1}^nfrac{k^2}{nk+k^2}leq1-frac{left(sumlimits_{k=1}^nkright)^2}{nsumlimits_{k=1}^n(nk+k^2)}=1-frac{frac{n(n+1)^2}{4}}{ncdotfrac{n(n+1)}{2}+frac{n(n+1)(2n+1)}{6}}=$$
      $$=frac{7n-1}{2(5n+1)}<0.7leqfrac{3}{4}.$$
      Done!



      I think it's interesting that $ln2=0.6931...$.



      C-S forever!!!






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Interesting is that $frac{7n-1}{2(5n+1)}=frac 7 {10}-frac 6{25n}+Oleft(frac{1}{n^2}right)$ to compare with $S_n=log (2)-frac{1}{4 n}+Oleft(frac{1}{n^2}right)$ I gave in a deleted answer based on the asymptotics. Look how close to each other are the coefficients. By the way, $to +1$.
        $endgroup$
        – Claude Leibovici
        Aug 6 '17 at 14:54
















      4












      4








      4





      $begingroup$

      By C-S
      $$sum_{k=1}^nfrac{1}{n+k}=1-sum_{k=1}^nleft(frac{1}{n}-frac{1}{n+k}right)=1-frac{1}{n}sum_{k=1}^nfrac{k}{n+k}=$$
      $$=1-frac{1}{n}sum_{k=1}^nfrac{k^2}{nk+k^2}leq1-frac{left(sumlimits_{k=1}^nkright)^2}{nsumlimits_{k=1}^n(nk+k^2)}=1-frac{frac{n(n+1)^2}{4}}{ncdotfrac{n(n+1)}{2}+frac{n(n+1)(2n+1)}{6}}=$$
      $$=frac{7n-1}{2(5n+1)}<0.7leqfrac{3}{4}.$$
      Done!



      I think it's interesting that $ln2=0.6931...$.



      C-S forever!!!






      share|cite|improve this answer









      $endgroup$



      By C-S
      $$sum_{k=1}^nfrac{1}{n+k}=1-sum_{k=1}^nleft(frac{1}{n}-frac{1}{n+k}right)=1-frac{1}{n}sum_{k=1}^nfrac{k}{n+k}=$$
      $$=1-frac{1}{n}sum_{k=1}^nfrac{k^2}{nk+k^2}leq1-frac{left(sumlimits_{k=1}^nkright)^2}{nsumlimits_{k=1}^n(nk+k^2)}=1-frac{frac{n(n+1)^2}{4}}{ncdotfrac{n(n+1)}{2}+frac{n(n+1)(2n+1)}{6}}=$$
      $$=frac{7n-1}{2(5n+1)}<0.7leqfrac{3}{4}.$$
      Done!



      I think it's interesting that $ln2=0.6931...$.



      C-S forever!!!







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Aug 6 '17 at 10:14









      Michael RozenbergMichael Rozenberg

      111k1897201




      111k1897201












      • $begingroup$
        Interesting is that $frac{7n-1}{2(5n+1)}=frac 7 {10}-frac 6{25n}+Oleft(frac{1}{n^2}right)$ to compare with $S_n=log (2)-frac{1}{4 n}+Oleft(frac{1}{n^2}right)$ I gave in a deleted answer based on the asymptotics. Look how close to each other are the coefficients. By the way, $to +1$.
        $endgroup$
        – Claude Leibovici
        Aug 6 '17 at 14:54




















      • $begingroup$
        Interesting is that $frac{7n-1}{2(5n+1)}=frac 7 {10}-frac 6{25n}+Oleft(frac{1}{n^2}right)$ to compare with $S_n=log (2)-frac{1}{4 n}+Oleft(frac{1}{n^2}right)$ I gave in a deleted answer based on the asymptotics. Look how close to each other are the coefficients. By the way, $to +1$.
        $endgroup$
        – Claude Leibovici
        Aug 6 '17 at 14:54


















      $begingroup$
      Interesting is that $frac{7n-1}{2(5n+1)}=frac 7 {10}-frac 6{25n}+Oleft(frac{1}{n^2}right)$ to compare with $S_n=log (2)-frac{1}{4 n}+Oleft(frac{1}{n^2}right)$ I gave in a deleted answer based on the asymptotics. Look how close to each other are the coefficients. By the way, $to +1$.
      $endgroup$
      – Claude Leibovici
      Aug 6 '17 at 14:54






      $begingroup$
      Interesting is that $frac{7n-1}{2(5n+1)}=frac 7 {10}-frac 6{25n}+Oleft(frac{1}{n^2}right)$ to compare with $S_n=log (2)-frac{1}{4 n}+Oleft(frac{1}{n^2}right)$ I gave in a deleted answer based on the asymptotics. Look how close to each other are the coefficients. By the way, $to +1$.
      $endgroup$
      – Claude Leibovici
      Aug 6 '17 at 14:54













      2












      $begingroup$

      $$sum_{k=1}^{n}frac{1}{n+k}leq sum_{k=1}^{n}frac{1}{sqrt{(n+k)(n+k-1)}}stackrel{text{CS}}{leq}sqrt{nsum_{k=1}^{n}left(frac{1}{n+k-1}-frac{1}{n+k}right)}$$
      immediately leads to $H_{2n}-H_n = sum_{k=1}^{n}frac{1}{n+k}leq frac{1}{sqrt{2}}<frac{3}{4}$.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        $$sum_{k=1}^{n}frac{1}{n+k}leq sum_{k=1}^{n}frac{1}{sqrt{(n+k)(n+k-1)}}stackrel{text{CS}}{leq}sqrt{nsum_{k=1}^{n}left(frac{1}{n+k-1}-frac{1}{n+k}right)}$$
        immediately leads to $H_{2n}-H_n = sum_{k=1}^{n}frac{1}{n+k}leq frac{1}{sqrt{2}}<frac{3}{4}$.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          $$sum_{k=1}^{n}frac{1}{n+k}leq sum_{k=1}^{n}frac{1}{sqrt{(n+k)(n+k-1)}}stackrel{text{CS}}{leq}sqrt{nsum_{k=1}^{n}left(frac{1}{n+k-1}-frac{1}{n+k}right)}$$
          immediately leads to $H_{2n}-H_n = sum_{k=1}^{n}frac{1}{n+k}leq frac{1}{sqrt{2}}<frac{3}{4}$.






          share|cite|improve this answer









          $endgroup$



          $$sum_{k=1}^{n}frac{1}{n+k}leq sum_{k=1}^{n}frac{1}{sqrt{(n+k)(n+k-1)}}stackrel{text{CS}}{leq}sqrt{nsum_{k=1}^{n}left(frac{1}{n+k-1}-frac{1}{n+k}right)}$$
          immediately leads to $H_{2n}-H_n = sum_{k=1}^{n}frac{1}{n+k}leq frac{1}{sqrt{2}}<frac{3}{4}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 6 '17 at 15:37









          Jack D'AurizioJack D'Aurizio

          292k33284674




          292k33284674























              1












              $begingroup$

              All you really need to do is change the lower limit in the integral from $1$ to $0$, which gives the simpler result



              $$sum_{k=1}^n{1over n+k}leint_0^n{dxover n+x}=ln(2n)-ln n=ln2$$



              One way to see that $1$ is the wrong lower limit to use is that, in general, the number of terms in the sum should equal the difference of the upper and lower limits in the integral. That's because the integral comparison test usually compares each term with the area beneath a curve over a unit segment.






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              $endgroup$


















                1












                $begingroup$

                All you really need to do is change the lower limit in the integral from $1$ to $0$, which gives the simpler result



                $$sum_{k=1}^n{1over n+k}leint_0^n{dxover n+x}=ln(2n)-ln n=ln2$$



                One way to see that $1$ is the wrong lower limit to use is that, in general, the number of terms in the sum should equal the difference of the upper and lower limits in the integral. That's because the integral comparison test usually compares each term with the area beneath a curve over a unit segment.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  All you really need to do is change the lower limit in the integral from $1$ to $0$, which gives the simpler result



                  $$sum_{k=1}^n{1over n+k}leint_0^n{dxover n+x}=ln(2n)-ln n=ln2$$



                  One way to see that $1$ is the wrong lower limit to use is that, in general, the number of terms in the sum should equal the difference of the upper and lower limits in the integral. That's because the integral comparison test usually compares each term with the area beneath a curve over a unit segment.






                  share|cite|improve this answer









                  $endgroup$



                  All you really need to do is change the lower limit in the integral from $1$ to $0$, which gives the simpler result



                  $$sum_{k=1}^n{1over n+k}leint_0^n{dxover n+x}=ln(2n)-ln n=ln2$$



                  One way to see that $1$ is the wrong lower limit to use is that, in general, the number of terms in the sum should equal the difference of the upper and lower limits in the integral. That's because the integral comparison test usually compares each term with the area beneath a curve over a unit segment.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Aug 6 '17 at 11:35









                  Barry CipraBarry Cipra

                  60.8k655129




                  60.8k655129






























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