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infinite series 2

sbhatnagar

Active member
Jan 27, 2012
95
More fun problems! Evaluate the following:

1. \( \displaystyle \sum_{n=0}^{\infty}\frac{(-1)^n \{ (2n+1)\sin^{2n+1}(k)+(-1)^n\cos^{2n+1}(k) \}}{\cos^{2n+1}(k)(2n+1)^2} \)

2. \( \displaystyle \sum_{n=1}^{\infty} \frac{\sin^4(2^n)}{4^n} \)

3. \( \displaystyle \sum_{n=0}^{\infty}\frac{\{ n+(n-1)x^2\}^2}{n!}\cdot \frac{x^{2n}}{(1+x^2)^{n+2}}\cdot \exp{ \left( \frac{-x^2}{1+x^2}\right)}\)

4. \( \displaystyle \sum_{n=1}^{\infty}\frac{2}{(2n-1)(2x+1)^{2n-1}} \)
 
Jan 31, 2012
54
4.


Consider the series:


$$\sum_n^{\infty}(\frac{1}{2x+1})^2n=\frac{4x^2+4x}{(2x+1)^2}$$ (1)


Now, if we integrate (1), we will get:


$$T=\sum_n^{\infty}\int(\frac{1}{2x+1})^{2n} dx=\sum_n^{\infty}\frac{-1}{2(2n-1)(2x+1)^{2n-1}}$$


If $S$ our original sum, then:


$$S=-4T=-4\int\frac{4x^2+4x}{(2x+1)^2}dx=-4(x+\frac{2}{2x+1})$$
 

sbhatnagar

Active member
Jan 27, 2012
95
Hi also sprach zarathustra! You made a mistake.
4.


Consider the series:


$$\sum_{n=1}^{\infty}\left(\frac{1}{2x+1}\right)^{2n}={\color{red}{\frac{1}{4x(x+1)}}} $$ (1)


now, if we integrate (1), we will get:


$$t=\sum_{n=1}^{\infty}\int \left(\frac{1}{2x+1} \right)^{2n} dx=\sum_{n=1}^{\infty}\frac{-1}{2(2n-1)(2x+1)^{2n-1}}$$


if $s$ our original sum, then:


$$s=-4t=-4 \color{red}{\int\frac{1}{4x(x+1)}dx=-\int \frac{1}{x}-\frac{1}{x+1} dx = -\ln{x}+\ln{(x+1)}=\ln{\left( \frac{x+1}{x}\right)} }$$
My approach was quite similar to yours except that I directly used the expansion of ln(1+x).

\( \displaystyle \ln{\left(1+\frac{1}{2x+1} \right)}=-\sum_{n=1}^{\infty}\frac{(-1)^n }{n(2x+1)^n} \quad (1)\)

\( \displaystyle \ln{\left(1-\frac{1}{2x+1} \right)}=-\sum_{n=1}^{\infty}\frac{1}{n(2x+1)^n} \quad (2)\)

Subtracting (2) from (1):

\( \displaystyle \ln{\left(1+\frac{1}{2x+1} \right)}-\ln{\left(1-\frac{1}{2x+1} \right)}= \sum_{0}^{\infty}\frac{2}{(2n-1)(2x+1)^{2n-1}}\)

\(\Rightarrow \displaystyle \ln{\left(\frac{x+1}{x} \right)}= \sum_{0}^{\infty}\frac{2}{(2n-1)(2x+1)^{2n-1}}\)

\(\displaystyle \Rightarrow \sum_{0}^{\infty}\frac{2}{(2n-1)(2x+1)^{2n-1}} =\boxed{2 \cdot\coth^{-1}(2x+1)} \)
 
Last edited:

sbhatnagar

Active member
Jan 27, 2012
95
It doesn't look like anybody else is going to post a solution.

\( \displaystyle \sum_{n=0}^{\infty}\frac{(-1)^n \{ (2n+1)\sin^{2n+1}(k)+(-1)^n\cos^{2n+1}(k) \}}{\cos^{2n+1}(k)(2n+1)^2} \)

Let \( \displaystyle S=\sum_{n=0}^{\infty}\frac{(-1)^n \{ (2n+1)\sin^{2n+1}(k)+(-1)^n\cos^{2n+1}(k) \}}{\cos^{2n+1}(k)(2n+1)^2} \)

\( \displaystyle \Rightarrow S=\sum_{n=0}^{\infty}\frac{(-1)^n (2n+1)\tan^{2n+1}(k)+1}{(2n+1)^2} \)

\( \displaystyle \Rightarrow S=\sum_{n=0}^{\infty}\left[\frac{(-1)^n \tan^{2n+1}(k)}{2n+1} +\frac{1}{(2n+1)^2} \right] \)

It is known that \( \displaystyle \arctan(z)=\sum_{n=0}^{\infty}\frac{(-1)^n z^{2n+1}}{2n+1} \), so we get:

\( \displaystyle S= \arctan(\tan(k))+\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2} \)

\( \displaystyle \Rightarrow S= k+\left( \frac{\pi^2}{6}-\frac{1}{4}\frac{\pi^2}{6}\right) \)

\( \displaystyle \Rightarrow S= \boxed{\displaystyle k+\frac{\pi^2}{8}} \)


\( \displaystyle \sum_{n=1}^{\infty} \frac{\sin^4(2^n)}{4^n} \)

Note That: \(\displaystyle \frac{\sin^4(2^n)}{4^n}=\frac{\sin^2(2^n) \cdot \sin^2(2^n)}{4^n}= \frac{\sin^2(2^n) \cdot (1-\cos^2(2^n))}{4^n}= \frac{\sin^2(2^n)}{4^n}-\frac{\sin^2(2^{n+1})}{4^{n+1}}\)

\( \displaystyle \sum_{n=1}^{\infty} \frac{\sin^4(2^n)}{4^n} = \sum_{n=1}^{\infty} \frac{\sin^2(2^n)}{4^n}-\frac{\sin^2(2^{n+1})}{4^{n+1}}=\frac{\sin^2(2)}{4} +\sum_{n=2}^{\infty} \frac{\sin^2(2^n)}{4^n}-\sum_{n=1}^{\infty}\frac{\sin^2(2^{n+1})}{4^{n+1}} = \frac{\sin^2(2)}{4}+0=\boxed{\displaystyle \frac{\sin^2(2)}{4}}\)


\( \displaystyle \sum_{n=0}^{\infty}\frac{\{ n+(n-1)x^2\}^2}{n!}\cdot \frac{x^{2n}}{(1+x^2)^{n+2}}\cdot \exp{ \left( \frac{-x^2}{1+x^2}\right)}\)

Let \( \displaystyle S=\sum_{n=0}^{\infty}\frac{\{ n+(n-1)x^2\}^2}{n!}\cdot \frac{x^{2n}}{(1+x^2)^{n+2}}\cdot \exp{ \left( \frac{-x^2}{1+x^2}\right)}\)

\( \displaystyle \Rightarrow S=\sum_{n=0}^{\infty}\frac{\left\{ n - \frac{x^2}{1+x^2} \right\}^2}{n!}\cdot \left(\frac{x^{2}}{1+x^2} \right)^n\cdot \exp{ \left( \frac{-x^2}{1+x^2}\right)}\)

Substitute \( \displaystyle y=\frac{x^2}{1+x^2} \).

\( \displaystyle \begin{align*} S &=e^{-y}\sum_{n=0}^{\infty}\frac{y^n(n-y)^2}{n!} \\ &=e^{-y}\sum_{n=0}^{\infty}\frac{n^2 y^n+y^{n+2}-2n y^{n+1}}{n!} \\ &= e^{-y} \left[\sum_{n=0}^{\infty}\frac{n^2 y^n}{n!}+ \sum_{n=0}^{\infty}\frac{ y^{n+2}}{n!}-\sum_{n=0}^{\infty}\frac{2n y^{n+1}}{n!}\right] \\ &= e^{-y} \left[y\sum_{n=0}^{\infty}\frac{n y^{n-1}}{(n-1)!}+ y^2\sum_{n=0}^{\infty}\frac{ y^{n}}{n!}-2y^2\sum_{n=0}^{\infty}\frac{ y^{n-1}}{(n-1)!}\right] \\ &= e^{-y} \left[y^2\sum_{n=0}^{\infty}\frac{ y^{n-2}}{(n-2)!}+y\sum_{n=0}^{\infty}\frac{ y^{n-1}}{(n-1)!}+ y^2e^y-2y^2 e^y\right]\\ &= e^{-y}(y e^y +2y^2 e^y-2y^2 e^y) \\&= e^{-y}(y e^y) \\&= y \end{align*}\)

\( \displaystyle S=y=\boxed{\displaystyle \frac{x^2}{1+x^2}} \).
 

oasi

New member
Mar 14, 2012
14
More fun problems! Evaluate the following:

1. \( \displaystyle \sum_{n=0}^{\infty}\frac{(-1)^n \{ (2n+1)\sin^{2n+1}(k)+(-1)^n\cos^{2n+1}(k) \}}{\cos^{2n+1}(k)(2n+1)^2} \)

2. \( \displaystyle \sum_{n=1}^{\infty} \frac{\sin^4(2^n)}{4^n} \)

3. \( \displaystyle \sum_{n=0}^{\infty}\frac{\{ n+(n-1)x^2\}^2}{n!}\cdot \frac{x^{2n}}{(1+x^2)^{n+2}}\cdot \exp{ \left( \frac{-x^2}{1+x^2}\right)}\)

4. \( \displaystyle \sum_{n=1}^{\infty}\frac{2}{(2n-1)(2x+1)^{2n-1}} \)
our teacher asked to us 4. problem