How to get sum of $\frac{1}{1+x^2}+\frac{1}{(1+x^2)^2}+...+\frac{1}{(1+x^2)^n}$ using mathematical induction

Question

Prehistory: I'm reading book. Because of exercises, reading process is going very slowly. Anyway, I want honestly complete all exercises. Theme in the book is mathematical induction. There were examples, where were shown how with mathematical induction prove equations like $(1+q)(1+q^2)(1+q^4)\dots(1+q^{{2}^{n}}) = \frac{1-q^{{2}^{n+1}}}{1-q}$.

Now I'm trying to complete exercise where I have to find sum of $\frac{1}{1+x^2}+\frac{1}{(1+x^2)^2}+...+\frac{1}{(1+x^2)^n}$. I tried to do it with mathematical induction. Like this: n=1: $\frac{1}{1+x^2}$

n=2: $\frac{1}{1+x^2}+\frac{1}{(1+x^2)^2}=\frac{1+1+x^2}{(1+x^2)^2}$

n=3: $\frac{1}{1+x^2}+\frac{1}{(1+x^2)^2}+\frac{1}{(1+x^2)^3}=\frac{(1+x^2)^2+2+x^2}{(1+x^2)^3}$

...

And so on (I've calculated til n=5). But I don't see any consistent pattern to evaluate sum of progression.

After that I found formulas of geometrical progression: $q=\frac{b_{n+1}}{b_n}$ and $S_n=\frac{b_1(1-q^n)}{1-q}$, so I've evaluated: $q=\frac{\frac{1}{(1+x^2)^2}}{\frac{1}{1+x^2}}=\frac{1}{1+x^2}$

and

$S_n=\left(\frac{1}{1+x^2}\left[1-\left(\frac{1}{1+X^2}\right)^n\right]\right):\left(1-\left[\frac{1}{1+x^2}\right]\right)=\left[\frac{1}{1+x^2}-\left(\frac{1}{1+x^2}\right)^{n+1}\right]\frac{1+x^2}{x^2}=\frac{\left[1-\left(\frac{\sqrt[n+1]{1+x^2}}{1+x^2}\right)^{n+1}\right]}{x^2}$

First of all, I'd like to know how to find sum of geometrical progression with mathematical induction. Secondly, I'd like to know what is wrong with my evaluations.

Answer 1

For any $n$ in your calculations(which are right but missed the right trick) use the following:

first set $${{S}_{n}}=q+{{q}^{2}}+\ldots +{{q}^{n}}$$ then $$q{{S}_{n}}={{q}^{2}}+{{q}^{3}}+\ldots +{{q}^{n+1}}$$ al last $$\begin{align} & {{S}_{n}}-q{{S}_{n}}=q-{{q}^{n+1}} \\ & {{S}_{n}}=\frac{q\left( 1-{{q}^{n}} \right)}{\left( 1-q \right)} \\ \end{align}$$

Answer 2

Hint:

For a geometric series with ratio $q$: $$q+q^2+\dots+q^n=\frac{q(1-q^n)}{1-q}.$$

Answer 3

You like to prove $$q+q^2+\dots+q^n=\frac{q(1-q^n)}{1-q}.$$ using Mathematical induction.

The expression is true for $n=1$ because it is simply $q=q$

If true for n, you want to show that it is also true for $n+1$

Note that $$ q+q^2+\dots+q^n+q^{n+1} = \frac{q(1-q^n)}{1-q} + q^{n+1} = $$

$$ \frac {q(1-q^n)+(1-q)q^{n+1}}{1-q}= \frac {q(1-q^{n+1})}{1-q} $$

With $q= \frac {1}{1+x^2}$ the expression under question simplifies to $$\frac{1}{1+x^2}+\frac{1}{(1+x^2)^2}+…+\frac{1}{(1+x^2)^n}=\frac {(1+x^2)^n-1}{x^2(1+x^2)^n}$$