Q:

How to prove that series $\frac{1}{n+1}$, as $n\to \infty$ is zero.

Can somebody explain how to prove that series $\frac{1}{n+1}$, as $n \to \infty$? I mean infinite series, not sequence, and I want to understand how to define the partial sum when n goes to infinity. Thank you!

A:

The natural numbers are not bounded (this can be proven by using the fact the real numbers have the least upper bound property). What this means is given a real number $r$ we can find a natural number $N$ so that $N>r$.
We wish to prove $\lim_{n\to\infty}\frac{1}{n+1}=0$. So we have to prove given $\epsilon>0$ there is an $N$ such that if $n\geq N$ then $|0-\frac{1}{n+1}|<\epsilon$ so $\frac{1}{n+1}<\epsilon$
Of course this is equivalent to $n+1>\frac{1}{\epsilon}\iff n>\frac{1}{\epsilon}-1$. Because the natural numbers are not bounded, there is an $N$ that satisfies that inequality. This is the $N$ we need.
Because from $n\geq N$ we have $n>\frac{1}{\epsilon}-1$ and this gives us $\frac{1}{n+1}<\epsilon$ which is what we needed.

You changed the problem statement, apparently you want to prove $\lim_{n\to\infty}\sum\limits_{i=1}^n\frac{1}{i+1}=0$. This is not true, a very simple way to see this is that every term of the series is larger than or equal to $1$, hence given $\epsilon<1$ it will be impossible to find $N$ so that if $n\geq N$ then $|0-\sum\limits_{i=1}^n\frac{1}{i+1}|<\epsilon$. Since we shall have $|0-\sum\limits_{i=1}^n\frac{1}{i+1}|\geq 1$.
It can also be proven that $\lim_{n\to\infty}\sum\limits_{i=1}^n\frac{1}{i+1}$ does not exist. I can write the proof if you want.

