<s>[INST] How to solve the recurrence $T(n)=T(n-1)+\frac{n}{n-1}$? 
Solve the recurrence $T(n)=T(n-1)+\frac{n}{n-1}$.

I thought to try solving it by iteration method:
$$
T(n)=\frac{n}{n-1}+\frac{n-1}{n-2}+\frac{n-2}{n-3}+...+\frac{3}{2}+\frac{2}{1}=\sum_{i=0}^n \frac{i}{i-1}
$$
 I know that $\sum^n_i i=0.5n(n+1)$ but I don't know how to develop the series from here.
 [/INST] Well, at first let $T(1)=a\in\mathbb{R}$ be the first term of the sequence $T(n)$. Then, let us try to find some of the next terms of the sequence:
$$T(2)=T(1)+\frac{2}{2-1}=a+2$$
$$T(3)=T(2)+\frac{3}{3-1}=a+2+\frac{3}{2}=a+\frac{7}{2}(=a+3.5)$$
$$T(4)=T(3)+\frac{4}{4-1}=a+\frac{7}{2}+\frac{4}{3}=a+\frac{29}{6}(=a+4.8\overline{3})$$
And, in general:
$$T(n)=a+\sum_{k=2}^n\frac{k}{k-1}$$
Now, we have:
$$\sum_{k=2}^n\frac{k}{k-1}=\sum_{k=2}^n\frac{k-1+1}{k-1}=\sum_{k=2}^n\left(1+\frac{1}{k-1}\right)=n-1+\sum_{k=2}^n\frac{1}{k-1}$$
So, we have to calculate the sum:
$$\sum_{k=2}^n\frac{1}{k-1}=\sum_{k=1}^{n-1}\frac{1}{k}=H_{n-1}$$
Where $H_n$ is the $n$-th harmonic number.
So, we finally have:
$$T(n)=a+n-1+H_{n-1}$$
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