Why is $\lim\limits_{n\to\infty} \sum\limits_{k=1}^n \frac{1}{k} - \log (n) = \sum\limits_{k=1}^{\infty} \frac{1}{k}-\log(\frac{k+1}{k})$ So i try to get from $\lim\limits_{n\to\infty} \sum\limits_{k=1}^n \frac{1}{k} - \log (n)$ this $  \sum\limits_{k=1}^{\infty} \frac{1}{k}-\log(\frac{k+1}{k})$
 $\lim\limits_{n\to\infty} \sum\limits_{k=1}^n \frac{1}{k} - \log (n) \stackrel{(1)}{=}  \lim\limits_{n\to\infty} \sum\limits_{k=1}^n \frac{1}{k} + \sum\limits_{k=1}^{n-1} \log(k)-\log(k+1) \stackrel{(2)}{=} \lim\limits_{n\to\infty} \frac{1}{n} + \sum\limits_{k=1}^{n-1} \log(\frac{k}{k+1})+\frac{1}{k}$   
$ \stackrel{(3)}{=} \sum\limits_{k=1}^{\infty} \frac{1}{k}-\log(\frac{k+1}{k})$  
So in Step (1) i understand that i can pull apart the sum but i don't see why i can rewrite the log like this? In from Step (2) to (3) It's clear that  $\lim\limits_{n\to\infty} \frac{1}{n} = 0 $, but why is there a change inside the sum?
 A: (1) $$\sum_{k=1}^{n-1} (\log (k) - \log (k+1)) \\= (\log (n-1) - \log (n)) + (\log (n-2) - \log (n-1)) + \ldots + (\log (2) - \log (3)) +  (\underbrace{\log (1)}_{=0} - \log (2)) \\ =  -\log (n) + \underbrace{\log (n-1) -\log(n-1)}_{=0} + \underbrace{\log(n-2) - \ldots +- \log(3)}_{=0} + \underbrace{\log(2) - \log(2)}_{=0} \\ = - \log(n)$$
(2) $$\log(\frac k {k+1}) = \log(k) - \log(k+1) = -(\log(k+1) - \log(k)) = - \log(\frac {k+1} k)$$
A: Because  By telescopic sum we have $$\sum\limits_{k=1}^{n-1} \log(\frac{k+1}{k}) =\sum\limits_{k=1}^{n-1}\log(k+1)-\log(k) =\log (n)$$
Hence
$$\lim\limits_{n\to\infty} \sum\limits_{k=1}^{n-1} \frac{1}{k} - \log (n) = \lim\limits_{n\to\infty}\sum\limits_{k=1}^{n-1}\left( \frac{1}{k}-\log(\frac{k+1}{k})\right)= \sum\limits_{k=1}^{\infty} \left(\frac{1}{k}-\log(\frac{k+1}{k})\right)$$
A: Note that
$$\sum_{k=1}^n\left(\frac1k-\log\frac{k+1}k\right)
=\sum_{k=1}^n\frac1k-\sum_{k=1}^n(\log(k+1)-\log k)
=\sum_{k=1}^n\frac1k-\log(n+1)$$
(sum telescopes). So
$$\sum_{k=1}^\infty\left(\frac1k-\log\frac{k+1}k\right)
=\lim_{n\to\infty}\left[\sum_{k=1}^n\frac1k-\log(n+1)\right].$$
I'll leave it to you to check this equals
$$\lim_{n\to\infty}\left[\sum_{k=1}^n\frac1k-\log n\right].$$
