Induction proof of a series

Question

Suppose we have the series $$f(n) = \sum_{i=1}^n \frac{2i-1}{i^4 - 2 i^3 + 3 i^2 - 2 i + 2}.$$ As a hint it was said that this series "telescopes". I observed the pattern to be $f(n)=\frac{n^2}{n^2 +1}$. I wish to prove this via induction. The base case holds, since $f(1)=\frac{1}{2}$ which comes out of both the definition of $f(n)$ and my proposed formula. Now I wish to prove that for some arbitrary $k$ we assume $f(k)=\frac{k^2}{k^2+1}$, then I wish to prove that $f(k+1)=\frac{(k+1)^2}{(k+1)^2+1} (=\frac{k^2 +2k +1}{k^2 +2k +2})$.

I wanted to start out by noticing: $$f(k+1) = \sum_{i=1}^{k+1} \frac{2i-1}{i^4 - 2 i^3 + 3 i^2 - 2 i + 2}= \\f (k)+ \frac{2(k+1)-1}{(k+1)^4 - 2 (k+1)^3 + 3 (k+1)^2 - 2 (k+1) + 2}$$ This eventually simplifies to $$ f(n+1)= \frac{k^2}{k^2 +1} + \frac{2k+1}{k^4 +2k^3+3k^2+2k+2}$$

Answer 1

Note that$$\frac{i^2}{i^2+1}-\frac{(i-1)^2}{(i-1)^2+1}=\frac{2i-1}{i_4-2i^3+3i^2-2i+2}$$and that therefore\begin{align}\sum_{i=1}^n\frac{2i-1}{i^4-2i^3+3i^2-2i+2}&=\sum_{i=1}^n\frac{i^2}{i^2+1}-\frac{(i-1)^2}{(i-1)^2+1}\\&=\frac{n^2}{n^2+1}-\frac{0^2}{0^2+1}\\&=\frac{n^2}{n^2+1}.\end{align}

Answer 2

For completeness sake I want to finish my own proof, without telescoping and simply by induction as I intended to:

Now notice that: $$(k^2+1) (k^2 +2k +2) =k^4 +2k^3 +3k^2+2k +2$$

We can thus write: $$f(k+1)=\frac{k^2}{k^2 +1}+ \frac{2k+1}{(k^2+1)(k^2+2k+2)} = \\ \frac{k^2 (k^2+2k+2)+ (2k+1)}{(k^2+1)(k^2+2k+2)}=\frac{k^4+2k^3+2k^2+2k+1}{(k^2+1)(k^2+2k+2)}$$

Finally, observe that: $$(k^2+1)(k^2+2k+1)=k^4+2k^3+2k^2+2k+1 $$

Such that we can write:

$$ f(k+1)=\frac{(k^2+1)(k^2+2k+1)}{(k^2+1)(k^2+2k+2)}=\frac{k^2+2k+1}{k^2+2k+2}=\frac{(k+1)^2}{(k+1)^2+1}$$

If the formula holds for arbitrary k, it also holds for $k+1$, by the principle of mathematical induction, it holds for all $n$ $\square$.