<s>[INST] Induction proof of a series 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
$$ f(n+1)= \frac{k^2}{k^2 +1} + \frac{2k+1}{k^4 +2k^3+3k^2+2k+2}$$
 [/INST] 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 $n$ $\square$.
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