# How to evaluate this infinite series :$1+\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}+\dfrac{1}{11}+....$ [duplicate]

How to evaluate this infinite series :$$S=1+\dfrac{1}{3}-\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{9}+\dfrac{1}{11}+....=1+\sum_{k=0}^{\infty}\frac{2(-1)^k}{(4k+3)(4k+5)}$$

therfore: $$S-1=\sum_{k=0}^{\infty}\frac{2(-1)^k}{(4k+3)(4k+5)}=\sum_{k=1}^{\infty}\frac{2(-1)^{k+1}}{(4k-1)(4k+1)}=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{(4k-1)}-\frac{(-1)^{k+1}}{(4k+1)}$$

Therfore: $$S-1=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{(4k-1)}-\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{(4k+1)}$$

How can I calculate this?

• Use the identity $\frac1n=\int_0^1x^{n-1}~{\rm d}x$. This should convert it into a geometric series, which can be summed and then integrated. Commented May 15 at 12:58
• You're allowed to separate a series into two subseries if the subseries both converge absolutely. However, in some series, commuting the terms changes the sum. What theorem allows separating this series into two subseries? Commented May 15 at 13:25
• I have asked this question in $2019$, math.stackexchange.com/questions/3190908/… Commented May 15 at 13:38

We have $$\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{(4k-1)}=\sum_{k=1}^{\infty}\int_0^1 (-1)^{k+1}x^{4k-2}\,dx=\int_0^1 \left(\sum_{k=1}^{\infty}(-1)^{k+1}x^{4k-2}\right)\,dx=\int_{0}^1\dfrac{x^2}{1+x^4}\,dx,$$

where the change of order between sum and integral is justified by the Dominated Convergence Theorem.

Analogously, $$\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{(4k+1)}=\sum_{k=1}^{\infty}\int_0^1 (-1)^{k+1}x^{4k}\,dx=\int_0^1 \left(\sum_{k=1}^{\infty}(-1)^{k+1}x^{4k}\right)\,dx=\int_{0}^1\dfrac{x^4}{1+x^4}\,dx,$$

Thus, we have $$S=1+\int_{0}^1\dfrac{x^2-x^4}{1+x^4}\,dx$$

An antiderivative of the integrand can be found by the standard methods (long division, partial fractions,etc) and is equal to $$\dfrac{\arctan\left(\frac{2x+\sqrt{2}}{\sqrt{2}}\right)}{\sqrt{2}}+\dfrac{\arctan\left(\frac{2x-\sqrt{2}}{\sqrt{2}}\right)}{\sqrt{2}}-x$$

Therefore, $$\int_{0}^1\dfrac{x^2-x^4}{1+x^4}\,dx=\dfrac {\arctan(1+\sqrt{2})-\arctan(1-\sqrt{2})}{\sqrt{2}}-1$$

Using that $$\arctan(a)-\arctan(b)=\arctan\left(\dfrac{a-b}{1+ab}\right)$$ we get $$\int_{0}^1\dfrac{x^2-x^4}{1+x^4}\,dx=\dfrac{\pi}{2\sqrt{2}}-1$$

from where it follows that $$S=1+\left(\dfrac{\pi}{2\sqrt{2}}-1\right)=\dfrac{\pi}{2\sqrt{2}}$$

We could rewrite the summation as $$\int_0^1 dx + \int_0^1 x^2 dx - \int_0^1 x^4 dx -\int_0^1 x^6 dx + ...$$ This could be written as $$\int_0^1 1+x^2-x^4-x^6+x^8+x^{10}+... dx$$ This could be be simplified by writing these terms as a part of a GP. We could add and subtract 2 times all the negative parts of the integration. We get $$\int_0^1 1+x^2+x^4+x^6+... dx - (-2\int_0^1 x^4+x^6+x^{12}+x^{14}... dx)$$ And now we could write the first part of the integration as $$\int_0^1 \frac{1}{1-x^2} dx$$ and similarly we could simplify the final part of the summation by separating each power of consecutive two (sorry for such confusing language), so the final summation becomes $$\int_0^1 x^4+x^{12}+x^{20}+... dx + \int_0^1 x^6+x^{14}+x^{22}+... dx$$. This can now overall be simplified to $$\int_0^1 \frac{1}{1-x^2} dx + 2( \int_0^1 \frac{x^4}{1-x^8} dx + \int_0^1 \frac{x^6}{1-x^8} dx)$$. Now this question can be easily solved. Hope this helps.

• Hi, welcome to Math SE. Thanks for using MathJax to write your answer. Please note x^n only works as intended when the exponent is a single-character or a call to a backslashed method such as \frac{a}{b}. To get $x^{10}$, you need some curly braces viz.x^{10}.
– J.G.
Commented May 15 at 13:25
• The last integral is obviously a mistake, unless you meant 1/1. I'll leave it to you work out what you meant. perhaps \frac{1}{1-x^2} etc. Commented May 15 at 13:37