sequences-and-series
I came across these series while solving a probability question.
let |r| < 1 ,
$$S_n=\sum_{k=0}^{\infty}k^n.r^k$$
For n=0 ,it's a GP. $S_0=\frac{1}{1-r}$
For n=1 ,it's a AGP , $S_1=\frac{-1}{1-r}+\frac{1}{(1-r)^2}$
for n=2 , This series can be reduced to AGP by substituting $k^2=1.3.5….(2k-1)$ & $S_2=\frac{-r}{(1-r)^2}+\frac{2r}{(1-r)^3}$.
Is it possible to find sum further in this series . Is there any pattern.
The $S=\sum_{k=1}^\infty \frac{10}{10k+1}$ series is divergent. The parial sums have closed-form. Let denote
$$S_n = \sum_{k=1}^n \frac{10}{10k+1}.$$
Then in terms of digamma function
$$S_n = \psi\left(n+\frac{11}{10}\right) - \psi\left(\frac{1}{10}\right) - 10.$$
Here $\psi\left(\frac{1}{10}\right)$ has an elementary closed-form, but I don't know about an alternate form of the other $\psi$ term. The first few elements in the sequence are
$$0, \frac{10}{11},\frac{320}{231},\frac{12230}{7161},\frac{573040}{293601},\frac{3573450}{1663739},\frac{234617840}{101488079},\frac{17672747430}{7205653609},\dots$$
In general if the following sum exists, then:
$$\sum_{k=1}^n \frac{a}{bk+c} = \frac ab \left( \psi\left(\frac cb + n+1\right) - \psi\left(\frac cb + 1\right) \right).$$
To answer the modified question we could use direct comparison test. As @idm mentioned, since
$$\frac{10}{10n+1}\geq\frac{1}{2n} \geq 0,$$
and because $\sum \frac{1}{2n}$ diverges it follows that $\sum \frac{10}{10n+1}$ also diverges.
I can see two problems.
First off, your Fourier coefficients as given are only half as big as they should be to match your Series1 data. Perhaps you used the even nature of the target function so you only integrated over half the interval but then forgot to multiply by $2$ to take into account the half of the integral you didn't do.
Second, your value for $a_3$ is very high. I didn't actually look at your spreadsheet; here's my Matlab program:
% fouriertest.m
npts = 300;
x = linspace(-2,2,npts);
xgap = [-2 -1 -1 -1 0 0 0 1 1 1 2];
ygap = [20 20 NaN 10 0 NaN 0 10 NaN 20 20];
y0 = 25/2*ones(size(x));
y1 = (-20/pi-40/pi^2)*cos(pi/2*x);
y2 = -20/pi^2*cos(pi*x);
y3 = (20/3/pi-40/pi^2)*cos(3*pi/2*x);
plot(xgap,ygap,'b-',x,y0,'r-',x,y0+y1,'g-',x,y0+y1+y2,'m-', ...
x,y0+y1+y2+y3,'c-');
title('Partial Sums of Fourier Series');
xlabel('x');
ylabel('y');
legend('Series1','S_0','S_1','S_2','S_3','Location','southeast');
And its output:
BTW, your expression for the Fourier series has a typo which I an gonna fix.
Best Answer
If you take the (formal) dereivative of $S_n$ with respect to $r$, then $$ \frac{\mathrm d}{\mathrm dr}S_n=\sum_{k=1}^\infty k^n\frac{\mathrm d}{\mathrm dr}r^k=\sum_{k=1}^\infty k^n\cdot k r^{k-1}=\frac 1r\sum_{k=1}^ \infty k^{n+1}r^k=\frac1rS_{n+1}$$ so you obtain a recursion formula, $$S_{n+1}=r\frac{\mathrm d}{\mathrm dr}S_n. $$