Finding the sum of a numerical sequence

Question

Consider the series $\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)(2n+2)}$. Let S be its sum.

To solve this series i used the series of functions $\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+2}}{(2n+1)(2n+2)}$ of Sum S(x) and a radius of convergence R=1.

Now knowing the sum of $\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+2}}{(2n+1)(2n+2)}$ and then taking the limit : $\lim _ {x\rightarrow1}$S(x) would give us S.

Differentiating the Series of functions twice with respect to $x$ would lead us to a sum of a geomatric sequence $\sum_{n=0}^{\infty}(-1)^n x^{2n}= $=$1-x^2+x^4-x^6+...$! Can anyone help me in this from here with the integrating and getting the sum? since my integration didnt give me the true answer.

You are almost there. Note that the last sum is just $\frac{1}{1+x^2}$ — uniquesolution, Oct 17 '15 at 17:37

score 1 · Accepted Answer · answered Oct 17 '15 at 17:39

1

$$\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)(2n+2)}=\sum_{n\geq 0}(-1)^n\int_{0}^{1}\left(x^{2n}-x^{2n+1}\right)\,dx=\int_{0}^{1}\frac{1-x}{1+x^2}\,dx $$ hence:

$$\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)(2n+2)} = \color{red}{\frac{\pi}{4}-\frac{\log 2}{2}}.$$

answered Oct 17 '15 at 17:39

Jack D'Aurizio

353,855

Thankyou that's the true answer but can you tell me how you reached the integral part from the given series? – mandez Oct 17 '15 at 17:42
Ah found it thankyouu – mandez Oct 17 '15 at 17:43
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@MohamadMisto: I just exploited $\frac{1}{(2n+1)(2n+2)}=\frac{1}{2n+1}-\frac{1}{2n+2}$ and $\frac{1}{m+1}=\int_{0}^{1}x^m,dx$. – Jack D'Aurizio Oct 17 '15 at 17:43

graydad · Answer 2 · 2015-10-17T17:48:05.687

1

I think you are on the right track! We know $$\sum_{n = 0}^\infty (-x^2)^{n} = \frac{1}{1+x^2}$$ for $x$ where $|x|<1$, and integrating the sum twice gives $$\sum_{n=0}^\infty \frac{(-1)^nx^{2n+2}}{(2n+1)(2n+2)}+Cx +D$$ To pattern match this sum to your sum, we want $C = D = 0$. Now we want to integrate twice $\frac{1}{1+x^2}$. The first integral is clearly $\arctan(x)+C$ where we know $C$ must be zero. So the problem (if you approach it this way) boils down to solving $$\int \arctan(x) \text{d}x$$ You should be able to complete the second integral through integration by parts. When you're done simply deduce that your next integration constant is also equal to $0$ and plug in $x=1$. Or just evaluate your integral from $0$ to $1$ to avoid dealing with the constant.

edited Oct 17 '15 at 17:48

answered Oct 17 '15 at 17:43

graydad

14,077

Here was my problem in finding C couldnt find the way, do you have in mind a way to find it? – mandez Oct 17 '15 at 17:46
@MohamadMisto I edited my answer, do you buy the reasoning? – graydad Oct 17 '15 at 17:49
Yes yes well thankyou, its now to find the integral of arctanx thankyou – mandez Oct 17 '15 at 17:53
@MohamadMisto It's a pretty easy one actually! Just make sure to let $dv = 1$. Good luck :) – graydad Oct 17 '15 at 17:54

E.H.E · Answer 3 · 2015-10-17T18:02:33.667

0

$$\tan^{-1}(x)=\frac{(-1)^nx^{2n+1}}{2n+1}$$ $$\int_{0}^{1}\int_{0}^{x}\tan^{-1}(x)dxdx=\int_{0}^{1}\int_{0}^{x}\frac{(-1)^nx^{2n+1}}{2n+1}dxdx=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)(2n+2)}$$

edited Oct 17 '15 at 18:02

answered Oct 17 '15 at 17:55

E.H.E

23,280

Finding the sum of a numerical sequence

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