Method for determining convergence of sinus/cosinus series
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
111k7105186
asked Nov 30 '18 at 12:12
sydsyd
52
add a comment |
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
111k7105186
asked Nov 30 '18 at 12:12
sydsyd
52
add a comment |
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
111k7105186
asked Nov 30 '18 at 12:12
sydsyd
52
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
111k7105186
asked Nov 30 '18 at 12:12
sydsyd
52
edited Nov 30 '18 at 12:50
Arthur
111k7105186
asked Nov 30 '18 at 12:12
sydsyd
52
edited Nov 30 '18 at 12:50
Arthur
111k7105186
edited Nov 30 '18 at 12:50
Arthur
111k7105186
edited Nov 30 '18 at 12:50
Arthur
111k7105186
111k7105186
asked Nov 30 '18 at 12:12
sydsyd
52
asked Nov 30 '18 at 12:12
sydsyd
52
asked Nov 30 '18 at 12:12
sydsyd
52
52
add a comment |
add a comment |
1 Answer
1
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oldest
votes
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
18.4k21340
Note that the convergence follows by the Leibniz criterion. :-)
– 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
votes
active
oldest
votes
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
18.4k21340
Note that the convergence follows by the Leibniz criterion. :-)
– p4sch
Nov 30 '18 at 13:41
add a comment |
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
18.4k21340
Note that the convergence follows by the Leibniz criterion. :-)
– p4sch
Nov 30 '18 at 13:41
add a comment |
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
18.4k21340
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
18.4k21340
answered Nov 30 '18 at 12:53
B. GoddardB. Goddard
18.4k21340
answered Nov 30 '18 at 12:53
B. GoddardB. Goddard
18.4k21340
answered Nov 30 '18 at 12:53
B. GoddardB. Goddard
18.4k21340
18.4k21340
Note that the convergence follows by the Leibniz criterion. :-)
– p4sch
Nov 30 '18 at 13:41
add a comment |
Note that the convergence follows by the Leibniz criterion. :-)
– p4sch
Nov 30 '18 at 13:41
Note that the convergence follows by the Leibniz criterion. :-)
– p4sch
Nov 30 '18 at 13:41
Note that the convergence follows by the Leibniz criterion. :-)
– p4sch
Nov 30 '18 at 13:41
add a comment |
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