Ways of Turning Sequences into Infinite Series Is writing a sequence as a telescoping sum the only way to turn a sequence into an infinite series? In particular: 

Let $s_n = 1 +\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}- \ln(n+1), \ n \geq 1$. Convert this into an infinite series.

So let $a_n = s_n-s_{n-1}$. Then $a_n = \frac{1}{n}-\ln(n+1)+ \ln n$. This is equivalent to $$\sum_{n=1}^{\infty} \left(\frac{1}{n}- \ln \frac{n+1}{n} \right)$$
What does the inside term represent? 

It may be worth noting the following.
$$
s_n = \sum\limits_{k = 1}^n {\frac{1}{k}}  - \int_1^{n + 1} {\frac{1}{x}\,dx}  = \sum\limits_{k = 1}^n {\frac{1}{k}}  - \sum\limits_{k = 1}^n {\int_k^{k + 1} {\frac{1}{x}\,dx} }  = \sum\limits_{k = 1}^n {\bigg(\frac{1}{k} - \int_k^{k + 1} {\frac{1}{x}\,dx} \bigg)} .
$$
Hence $s_n  = \sum\nolimits_{k = 1}^n {a_k } $ with $a_n \,(= s_n - s_{n-1})$ given by 
$$
a_n = \frac{1}{n} - \int_n^{n + 1} {\frac{1}{x}\,dx}  = \frac{1}{n} - \ln \frac{{n + 1}}{n}.
$$
Now, from the simple fact that, for any $x > 0$,
$$
0 < x - \ln (1 + x) < x^2 
$$
it follows that, for any $n \in \mathbb{N}$,
$$
0 < \frac{1}{n} - \ln \bigg(1 + \frac{1}{n}\bigg) < \frac{1}{{n^2 }},
$$
so
$$
0 < a_n < \frac{1}{{n^2 }}.
$$
Hence
$$
\sum\limits_{n = 1}^\infty  {\bigg(\frac{1}{n} - \ln \frac{{n + 1}}{n}\bigg)}  = \sum\limits_{n = 1}^\infty  {a_n }  < \infty .
$$
