Find the value of the infinite series: $(1/3^2)+(1/3^3)+(1/3^4)+...+(1/5^2)+(1/5^3)+...+(1/7^2)+(1/7^3)+...+...$

Question

Essentially the summation of all the reciprocals of odd numbers and their integer powers(starting from 2). A close approximation could be {$1-\ln(2)$}.

My working was:

Consider the series $1/2^2+1/2^3+....+ 1/3^2+1/3^3+....$(for all integers).

I can segment each row as an infinite gp. First one being equal to $1/2=1/1*2$ $Second=1/6=1/2*3, third=1/12=1/3*4.....$

$Assume f(x)=x^2/2*1+x^3/2*3+..... $

$df(x)/d(x)=x+x^2/2+x^3/3...$ Which is $-ln(1-x)-x$

So $f(x)$=integral of $-x-ln(1-x)=ln(1-x)(1-x)+x-x^2/2+C$ $f(-1)=2ln2-3/2+C$=even gps positive and odd negative

$f(1)$= original series$=1/2+C$

$(f(1)-f(-1))/2=({(1/3^2)+(1/3^3)+(1/3^4)+...}+{(1/5^2)+(1/5^3)+....}+{...}+..)={1-ln(2)}??$

I'm not sure about this though

Also find the value of $(1/2^2)+(1/2^3)+(1/2^4)+...+(1/4^2)+(1/4^3)+...+(1/6^2)+(1/6^3)+...+...$

Answer 1

HINT: The problem can be reduced to the following:

$$\sum_{k=2}^{\infty}\frac{1}{(2k-1)(2k-2)}$$ by using the summation of geometric progressions to infinite terms.

HINT 2:

You can work out from the previous hint that the summation is equal to: (try using partial fraction decomposition)

$$\sum_{k=2}^{\infty}\frac{(-1)^k}{k}= 1-ln(2)$$

Problem 2: You can again follow the same steps as earlier to obtain that the summation is:

$$\sum_{k=1}^{\infty}\frac{1}{(2k-1)(2k)} = \sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = ln(2)$$

Answer 2

For $n\ge 1$, we have a geometric series $$\sum_{k=2}^\infty\frac{1}{(2n+1)^k}=\frac{\frac{1}{(2n+1)^2}}{1-\frac{1}{2n+1}}=\frac{1}{(2n+1)^2-(2n+1)}=\frac{1}{4n^2+2n}.$$ Taking the sum over $n\ge 1$ gives $$\sum_{n=1}^\infty\frac{1}{4n^2+2n}.$$ Now use partial fraction decomposition to rewrite $\frac{1}{4n^2+2n}$ and you will find that the above sum is (almost) a familiar series.