Finding limit of a (Laurent?) series
$begingroup$
I've been practicing series for my upcoming Calculus 1 exam, and I've stumbled upon this one:
$1 + frac{1}{1 + 2} + frac{1}{1 + 2 + 3} + ... + frac{1}{1 + 2 + 3 + ... + n}$
The task is to find the limit.
sequences-and-series limits
edited Feb 17 at 15:29
Max
566217
asked Feb 17 at 14:25
RoseRose
216
$endgroup$
add a comment |
$begingroup$
I've been practicing series for my upcoming Calculus 1 exam, and I've stumbled upon this one:
$1 + frac{1}{1 + 2} + frac{1}{1 + 2 + 3} + ... + frac{1}{1 + 2 + 3 + ... + n}$
The task is to find the limit.
sequences-and-series limits
edited Feb 17 at 15:29
Max
566217
asked Feb 17 at 14:25
RoseRose
216
$endgroup$
add a comment |
$begingroup$
I've been practicing series for my upcoming Calculus 1 exam, and I've stumbled upon this one:
$1 + frac{1}{1 + 2} + frac{1}{1 + 2 + 3} + ... + frac{1}{1 + 2 + 3 + ... + n}$
The task is to find the limit.
sequences-and-series limits
edited Feb 17 at 15:29
Max
566217
asked Feb 17 at 14:25
RoseRose
216
$endgroup$
I've been practicing series for my upcoming Calculus 1 exam, and I've stumbled upon this one:
$1 + frac{1}{1 + 2} + frac{1}{1 + 2 + 3} + ... + frac{1}{1 + 2 + 3 + ... + n}$
The task is to find the limit.
sequences-and-series limits
sequences-and-series limits
edited Feb 17 at 15:29
Max
566217
asked Feb 17 at 14:25
RoseRose
216
edited Feb 17 at 15:29
Max
566217
asked Feb 17 at 14:25
RoseRose
216
edited Feb 17 at 15:29
Max
566217
edited Feb 17 at 15:29
Max
566217
edited Feb 17 at 15:29
Max
566217
566217
asked Feb 17 at 14:25
RoseRose
216
asked Feb 17 at 14:25
RoseRose
216
asked Feb 17 at 14:25
RoseRose
216
216
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
We have $$1+2+...+k=frac{k(k+1)}2$$
So the sum you need to compute is
$$begin{split}
sum_{k=1}^n frac 2{k(k+1)} &= sum_{k=1}^n 2left ( frac 1 k - frac 1 { k+1} right )\
&=2-frac 2 {n+1}\
&=frac {2n} {n+1}
end{split}$$
Now you can take the limit.
edited Feb 17 at 19:03
RQM
1213
answered Feb 17 at 14:34
Stefan LafonStefan Lafon
2,23019
$endgroup$
add a comment |
$begingroup$
Hint: Recall the sum of an arithmetic series of consecutive numbers $1,2, cdots, n$:
$$1+2+cdots+n=frac{n(n+1)}{2}$$
Take reciprocal of it and deal with partial fraction, the terms will be eliminated.
$$
frac{2}{n(n+1)}=2(frac{1}{n}-frac{1}{n+1})$$
After that, take the limit.
answered Feb 17 at 14:34
weilam06weilam06
30412
$endgroup$
add a comment |
$begingroup$
Hint: Use the fact that $$1+2+...+n=frac{n(n+1)}{2}$$
The series then becomes $$2sumlimits_{n=1}^infty left(frac{1}{n} - frac{1}{n+1} right)$$ which is a telescoping series.
answered Feb 17 at 14:33
Haris GusicHaris Gusic
1,251117
$endgroup$
add a comment |
$begingroup$
As $1+2+ldots +n = frac{n}{2}(n+1)$ you are looking for the sum of the series
$S = sum_{n=1}^{infty} frac{1}{frac{k}{2}(k+1)} = 2 sum_{k=1}^{infty} frac{1}{k(k+1)}$.
Nos separating you get that $S_n = 2sum_{k=1}^{n} frac{1}{k} - frac{1}{k+1} = 2(1 - frac{1}{n+1})$. Now taking limit when $n rightarrow infty$ you get that $S=2$
answered Feb 17 at 14:36
JoseSquareJoseSquare
82012
$endgroup$
add a comment |
$begingroup$
$small{a_2=(1+2)^{-1}, a_3= (1+2+3)^{-1},....}$
$small{a_k=(1+2+...k)^{-1} = 2(k(k+1))^{-1}=2(1/k -1/(k+1))}$
Telescopic sum
$1+sum_{k=2}^{infty} a_k =?$
answered Feb 17 at 14:38
Peter SzilasPeter Szilas
11.4k2822
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We have $$1+2+...+k=frac{k(k+1)}2$$
So the sum you need to compute is
$$begin{split}
sum_{k=1}^n frac 2{k(k+1)} &= sum_{k=1}^n 2left ( frac 1 k - frac 1 { k+1} right )\
&=2-frac 2 {n+1}\
&=frac {2n} {n+1}
end{split}$$
Now you can take the limit.
edited Feb 17 at 19:03
RQM
1213
answered Feb 17 at 14:34
Stefan LafonStefan Lafon
2,23019
$endgroup$
add a comment |
$begingroup$
We have $$1+2+...+k=frac{k(k+1)}2$$
So the sum you need to compute is
$$begin{split}
sum_{k=1}^n frac 2{k(k+1)} &= sum_{k=1}^n 2left ( frac 1 k - frac 1 { k+1} right )\
&=2-frac 2 {n+1}\
&=frac {2n} {n+1}
end{split}$$
Now you can take the limit.
edited Feb 17 at 19:03
RQM
1213
answered Feb 17 at 14:34
Stefan LafonStefan Lafon
2,23019
$endgroup$
add a comment |
$begingroup$
We have $$1+2+...+k=frac{k(k+1)}2$$
So the sum you need to compute is
$$begin{split}
sum_{k=1}^n frac 2{k(k+1)} &= sum_{k=1}^n 2left ( frac 1 k - frac 1 { k+1} right )\
&=2-frac 2 {n+1}\
&=frac {2n} {n+1}
end{split}$$
Now you can take the limit.
edited Feb 17 at 19:03
RQM
1213
answered Feb 17 at 14:34
Stefan LafonStefan Lafon
2,23019
$endgroup$
We have $$1+2+...+k=frac{k(k+1)}2$$
So the sum you need to compute is
$$begin{split}
sum_{k=1}^n frac 2{k(k+1)} &= sum_{k=1}^n 2left ( frac 1 k - frac 1 { k+1} right )\
&=2-frac 2 {n+1}\
&=frac {2n} {n+1}
end{split}$$
Now you can take the limit.
edited Feb 17 at 19:03
RQM
1213
answered Feb 17 at 14:34
Stefan LafonStefan Lafon
2,23019
edited Feb 17 at 19:03
RQM
1213
edited Feb 17 at 19:03
RQM
1213
edited Feb 17 at 19:03
RQM
1213
1213
answered Feb 17 at 14:34
Stefan LafonStefan Lafon
2,23019
answered Feb 17 at 14:34
Stefan LafonStefan Lafon
2,23019
answered Feb 17 at 14:34
Stefan LafonStefan Lafon
2,23019
2,23019
add a comment |
add a comment |
$begingroup$
Hint: Recall the sum of an arithmetic series of consecutive numbers $1,2, cdots, n$:
$$1+2+cdots+n=frac{n(n+1)}{2}$$
Take reciprocal of it and deal with partial fraction, the terms will be eliminated.
$$
frac{2}{n(n+1)}=2(frac{1}{n}-frac{1}{n+1})$$
After that, take the limit.
answered Feb 17 at 14:34
weilam06weilam06
30412
$endgroup$
add a comment |
$begingroup$
Hint: Recall the sum of an arithmetic series of consecutive numbers $1,2, cdots, n$:
$$1+2+cdots+n=frac{n(n+1)}{2}$$
Take reciprocal of it and deal with partial fraction, the terms will be eliminated.
$$
frac{2}{n(n+1)}=2(frac{1}{n}-frac{1}{n+1})$$
After that, take the limit.
answered Feb 17 at 14:34
weilam06weilam06
30412
$endgroup$
add a comment |
$begingroup$
Hint: Recall the sum of an arithmetic series of consecutive numbers $1,2, cdots, n$:
$$1+2+cdots+n=frac{n(n+1)}{2}$$
Take reciprocal of it and deal with partial fraction, the terms will be eliminated.
$$
frac{2}{n(n+1)}=2(frac{1}{n}-frac{1}{n+1})$$
After that, take the limit.
answered Feb 17 at 14:34
weilam06weilam06
30412
$endgroup$
Hint: Recall the sum of an arithmetic series of consecutive numbers $1,2, cdots, n$:
$$1+2+cdots+n=frac{n(n+1)}{2}$$
Take reciprocal of it and deal with partial fraction, the terms will be eliminated.
$$
frac{2}{n(n+1)}=2(frac{1}{n}-frac{1}{n+1})$$
After that, take the limit.
answered Feb 17 at 14:34
weilam06weilam06
30412
answered Feb 17 at 14:34
weilam06weilam06
30412
answered Feb 17 at 14:34
weilam06weilam06
30412
answered Feb 17 at 14:34
weilam06weilam06
30412
30412
add a comment |
add a comment |
$begingroup$
Hint: Use the fact that $$1+2+...+n=frac{n(n+1)}{2}$$
The series then becomes $$2sumlimits_{n=1}^infty left(frac{1}{n} - frac{1}{n+1} right)$$ which is a telescoping series.
answered Feb 17 at 14:33
Haris GusicHaris Gusic
1,251117
$endgroup$
add a comment |
$begingroup$
Hint: Use the fact that $$1+2+...+n=frac{n(n+1)}{2}$$
The series then becomes $$2sumlimits_{n=1}^infty left(frac{1}{n} - frac{1}{n+1} right)$$ which is a telescoping series.
answered Feb 17 at 14:33
Haris GusicHaris Gusic
1,251117
$endgroup$
add a comment |
$begingroup$
Hint: Use the fact that $$1+2+...+n=frac{n(n+1)}{2}$$
The series then becomes $$2sumlimits_{n=1}^infty left(frac{1}{n} - frac{1}{n+1} right)$$ which is a telescoping series.
answered Feb 17 at 14:33
Haris GusicHaris Gusic
1,251117
$endgroup$
Hint: Use the fact that $$1+2+...+n=frac{n(n+1)}{2}$$
The series then becomes $$2sumlimits_{n=1}^infty left(frac{1}{n} - frac{1}{n+1} right)$$ which is a telescoping series.
answered Feb 17 at 14:33
Haris GusicHaris Gusic
1,251117
answered Feb 17 at 14:33
Haris GusicHaris Gusic
1,251117
answered Feb 17 at 14:33
Haris GusicHaris Gusic
1,251117
answered Feb 17 at 14:33
Haris GusicHaris Gusic
1,251117
1,251117
add a comment |
add a comment |
$begingroup$
As $1+2+ldots +n = frac{n}{2}(n+1)$ you are looking for the sum of the series
$S = sum_{n=1}^{infty} frac{1}{frac{k}{2}(k+1)} = 2 sum_{k=1}^{infty} frac{1}{k(k+1)}$.
Nos separating you get that $S_n = 2sum_{k=1}^{n} frac{1}{k} - frac{1}{k+1} = 2(1 - frac{1}{n+1})$. Now taking limit when $n rightarrow infty$ you get that $S=2$
answered Feb 17 at 14:36
JoseSquareJoseSquare
82012
$endgroup$
add a comment |
$begingroup$
As $1+2+ldots +n = frac{n}{2}(n+1)$ you are looking for the sum of the series
$S = sum_{n=1}^{infty} frac{1}{frac{k}{2}(k+1)} = 2 sum_{k=1}^{infty} frac{1}{k(k+1)}$.
Nos separating you get that $S_n = 2sum_{k=1}^{n} frac{1}{k} - frac{1}{k+1} = 2(1 - frac{1}{n+1})$. Now taking limit when $n rightarrow infty$ you get that $S=2$
answered Feb 17 at 14:36
JoseSquareJoseSquare
82012
$endgroup$
add a comment |
$begingroup$
As $1+2+ldots +n = frac{n}{2}(n+1)$ you are looking for the sum of the series
$S = sum_{n=1}^{infty} frac{1}{frac{k}{2}(k+1)} = 2 sum_{k=1}^{infty} frac{1}{k(k+1)}$.
Nos separating you get that $S_n = 2sum_{k=1}^{n} frac{1}{k} - frac{1}{k+1} = 2(1 - frac{1}{n+1})$. Now taking limit when $n rightarrow infty$ you get that $S=2$
answered Feb 17 at 14:36
JoseSquareJoseSquare
82012
$endgroup$
As $1+2+ldots +n = frac{n}{2}(n+1)$ you are looking for the sum of the series
$S = sum_{n=1}^{infty} frac{1}{frac{k}{2}(k+1)} = 2 sum_{k=1}^{infty} frac{1}{k(k+1)}$.
Nos separating you get that $S_n = 2sum_{k=1}^{n} frac{1}{k} - frac{1}{k+1} = 2(1 - frac{1}{n+1})$. Now taking limit when $n rightarrow infty$ you get that $S=2$
answered Feb 17 at 14:36
JoseSquareJoseSquare
82012
answered Feb 17 at 14:36
JoseSquareJoseSquare
82012
answered Feb 17 at 14:36
JoseSquareJoseSquare
82012
answered Feb 17 at 14:36
JoseSquareJoseSquare
82012
82012
add a comment |
add a comment |
$begingroup$
$small{a_2=(1+2)^{-1}, a_3= (1+2+3)^{-1},....}$
$small{a_k=(1+2+...k)^{-1} = 2(k(k+1))^{-1}=2(1/k -1/(k+1))}$
Telescopic sum
$1+sum_{k=2}^{infty} a_k =?$
answered Feb 17 at 14:38
Peter SzilasPeter Szilas
11.4k2822
$endgroup$
add a comment |
$begingroup$
$small{a_2=(1+2)^{-1}, a_3= (1+2+3)^{-1},....}$
$small{a_k=(1+2+...k)^{-1} = 2(k(k+1))^{-1}=2(1/k -1/(k+1))}$
Telescopic sum
$1+sum_{k=2}^{infty} a_k =?$
answered Feb 17 at 14:38
Peter SzilasPeter Szilas
11.4k2822
$endgroup$
add a comment |
$begingroup$
$small{a_2=(1+2)^{-1}, a_3= (1+2+3)^{-1},....}$
$small{a_k=(1+2+...k)^{-1} = 2(k(k+1))^{-1}=2(1/k -1/(k+1))}$
Telescopic sum
$1+sum_{k=2}^{infty} a_k =?$
answered Feb 17 at 14:38
Peter SzilasPeter Szilas
11.4k2822
$endgroup$
$small{a_2=(1+2)^{-1}, a_3= (1+2+3)^{-1},....}$
$small{a_k=(1+2+...k)^{-1} = 2(k(k+1))^{-1}=2(1/k -1/(k+1))}$
Telescopic sum
$1+sum_{k=2}^{infty} a_k =?$
answered Feb 17 at 14:38
Peter SzilasPeter Szilas
11.4k2822
edited Feb 17 at 16:50
answered Feb 17 at 14:38
Peter SzilasPeter Szilas
11.4k2822
answered Feb 17 at 14:38
Peter SzilasPeter Szilas
11.4k2822
answered Feb 17 at 14:38
Peter SzilasPeter Szilas
11.4k2822
11.4k2822
add a comment |
add a comment |
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