Method for determining convergence of sinus/cosinus series
$begingroup$
What's the method used for testing convergence of these type of series?
$$
sum_{n = 2}^inftysinfrac{(n^2-1)pi}{n}
$$
calculus sequences-and-series
edited Nov 30 '18 at 12:50
Arthur
115k7116198
asked Nov 30 '18 at 12:12
sydsyd
12
$endgroup$
add a comment |
$begingroup$
What's the method used for testing convergence of these type of series?
$$
sum_{n = 2}^inftysinfrac{(n^2-1)pi}{n}
$$
calculus sequences-and-series
edited Nov 30 '18 at 12:50
Arthur
115k7116198
asked Nov 30 '18 at 12:12
sydsyd
12
$endgroup$
add a comment |
$begingroup$
What's the method used for testing convergence of these type of series?
$$
sum_{n = 2}^inftysinfrac{(n^2-1)pi}{n}
$$
calculus sequences-and-series
edited Nov 30 '18 at 12:50
Arthur
115k7116198
asked Nov 30 '18 at 12:12
sydsyd
12
$endgroup$
What's the method used for testing convergence of these type of series?
$$
sum_{n = 2}^inftysinfrac{(n^2-1)pi}{n}
$$
calculus sequences-and-series
calculus sequences-and-series
edited Nov 30 '18 at 12:50
Arthur
115k7116198
asked Nov 30 '18 at 12:12
sydsyd
12
edited Nov 30 '18 at 12:50
Arthur
115k7116198
asked Nov 30 '18 at 12:12
sydsyd
12
edited Nov 30 '18 at 12:50
Arthur
115k7116198
edited Nov 30 '18 at 12:50
Arthur
115k7116198
edited Nov 30 '18 at 12:50
Arthur
115k7116198
115k7116198
asked Nov 30 '18 at 12:12
sydsyd
12
asked Nov 30 '18 at 12:12
sydsyd
12
asked Nov 30 '18 at 12:12
sydsyd
12
12
add a comment |
add a comment |
1 Answer
1
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votes
$begingroup$
You have
$$sinleft(frac{(n^2-1)pi}{n}right) = sinleft(npi -frac{pi}{n}right) $$
$$=sin(npi)cos(pi/n) - cos(npi)sin(pi/n)$$
$$= 0 +(-1)^{n+1}sin(pi/n).$$
So on one hand you have an alternating series and on the other hand you have something asymptotic to the harmonic series.
answered Nov 30 '18 at 12:53
B. GoddardB. Goddard
18.9k21440
$endgroup$
$begingroup$
Note that the convergence follows by the Leibniz criterion. :-)
$endgroup$
– p4sch
Nov 30 '18 at 13:41
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
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votes
$begingroup$
You have
$$sinleft(frac{(n^2-1)pi}{n}right) = sinleft(npi -frac{pi}{n}right) $$
$$=sin(npi)cos(pi/n) - cos(npi)sin(pi/n)$$
$$= 0 +(-1)^{n+1}sin(pi/n).$$
So on one hand you have an alternating series and on the other hand you have something asymptotic to the harmonic series.
answered Nov 30 '18 at 12:53
B. GoddardB. Goddard
18.9k21440
$endgroup$
$begingroup$
Note that the convergence follows by the Leibniz criterion. :-)
$endgroup$
– p4sch
Nov 30 '18 at 13:41
add a comment |
$begingroup$
You have
$$sinleft(frac{(n^2-1)pi}{n}right) = sinleft(npi -frac{pi}{n}right) $$
$$=sin(npi)cos(pi/n) - cos(npi)sin(pi/n)$$
$$= 0 +(-1)^{n+1}sin(pi/n).$$
So on one hand you have an alternating series and on the other hand you have something asymptotic to the harmonic series.
answered Nov 30 '18 at 12:53
B. GoddardB. Goddard
18.9k21440
$endgroup$
$begingroup$
Note that the convergence follows by the Leibniz criterion. :-)
$endgroup$
– p4sch
Nov 30 '18 at 13:41
add a comment |
$begingroup$
You have
$$sinleft(frac{(n^2-1)pi}{n}right) = sinleft(npi -frac{pi}{n}right) $$
$$=sin(npi)cos(pi/n) - cos(npi)sin(pi/n)$$
$$= 0 +(-1)^{n+1}sin(pi/n).$$
So on one hand you have an alternating series and on the other hand you have something asymptotic to the harmonic series.
answered Nov 30 '18 at 12:53
B. GoddardB. Goddard
18.9k21440
$endgroup$
You have
$$sinleft(frac{(n^2-1)pi}{n}right) = sinleft(npi -frac{pi}{n}right) $$
$$=sin(npi)cos(pi/n) - cos(npi)sin(pi/n)$$
$$= 0 +(-1)^{n+1}sin(pi/n).$$
So on one hand you have an alternating series and on the other hand you have something asymptotic to the harmonic series.
answered Nov 30 '18 at 12:53
B. GoddardB. Goddard
18.9k21440
answered Nov 30 '18 at 12:53
B. GoddardB. Goddard
18.9k21440
answered Nov 30 '18 at 12:53
B. GoddardB. Goddard
18.9k21440
answered Nov 30 '18 at 12:53
B. GoddardB. Goddard
18.9k21440
18.9k21440
$begingroup$
Note that the convergence follows by the Leibniz criterion. :-)
$endgroup$
– p4sch
Nov 30 '18 at 13:41
add a comment |
$begingroup$
Note that the convergence follows by the Leibniz criterion. :-)
$endgroup$
– p4sch
Nov 30 '18 at 13:41
$begingroup$
Note that the convergence follows by the Leibniz criterion. :-)
$endgroup$
– p4sch
Nov 30 '18 at 13:41
$begingroup$
Note that the convergence follows by the Leibniz criterion. :-)
$endgroup$
– p4sch
Nov 30 '18 at 13:41
add a comment |
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