Does the infinite series $1/(n+1)$ diverge? If so, how would one go about showing this? I'm playing around with comparison proofs and wondering if theres a way to show this either diverges or converges (due to it's very close relation to 1/n). Would it be correct to say it diverges due to the fact that as n approaches infinity, the +1 becomes negligible?
 A: You're right about the $+1$ becoming negligible as $n$ approaches infinity. There are a few good ways to see that this series diverges.
For one thing, it's actually the same series as $\sum \frac{1}{n}$, but with the first term missing. Since convergence or divergence has nothing to do with the initial terms, the two series must behave similarly.

Here's a way to use a direct comparison test:
We have $\frac{1}{n+1} > \frac{1}{2n}$ for all $n\geq 2$. This is good, because the series $\frac{1}{2n}$ is simply $\frac12$ times the series $\frac{1}{n}.$

The limit comparison test works very well. We just note that $\displaystyle\lim_{n\to\infty} \frac{1/(n+1)}{1/n}=1$, and therefore the two series either both converge or both diverge. Since $\frac{1}{n}$ diverges, then we're good

The integral test works, too. Calculate the integral: $\int_1^\infty \frac{dx}{x+1}$. Noting that it diverges, you can conclude that the corresponding series does, as well.
A: It's just the $1/n$ sequence with one term deleted, so divergence is retained.
A: Because $\sum_{n=1}^\infty\frac{1}{n}$ diverges, so does $-1+\sum_{n=1}^\infty\frac{1}{n}$.  But $\sum_{n=1}^\infty\frac{1}{n+1}=-1+\sum_{n=1}^\infty\frac{1}{n}$ so this sum diverges as well.
A: First way
Observe that
$$
\sum_{n=1}^{+\infty}\frac1{1+n}=\sum_{n=2}^{+\infty}\frac1{n}=+\infty
$$
Second way
Your observation about the presence of $+1$ is negligible when $n$ becomes huge is right: the formal way to use this is to observe that both
$$
\sum_{n=1}^{+\infty}\frac1{1+n}\;\;\;\mbox{and}\;\;\;
\sum_{n=1}^{+\infty}\frac1{n}
$$
have positive terms, and since
$$
\lim_n\frac{\frac1n}{\frac1{1+n}}=1
$$
the two series have the same behaviour; and since the latter one diverges, so does the first one.
Third way
\begin{align*}
\sum_{n=1}^{+\infty}\frac1{1+n}
&=\overbrace{\left(\frac12\right)}^{2^0 term}+\overbrace{\left(\frac13+\frac14\right)}^{2^1 terms}+
\overbrace{\left(\frac15+\frac16+\frac17+\frac18\right)}^{2^2 terms}+\\
&\cdots+
\underbrace{\left(\frac1{2^{k}+1}+\cdots+\frac1{2^{k+1}}\right)}_{2^k terms}+\cdots>\\
>&\overbrace{\left(\frac12\right)}^{2^0 term}+\overbrace{\left(\frac14+\frac14\right)}^{2^1 terms}+
\overbrace{\left(\frac18+\frac18+\frac18+\frac18\right)}^{2^2 terms}+\\
&\cdots+
\underbrace{\left(\frac1{2^{k+1}}+\cdots+\frac1{2^{k+1}}\right)}_{2^k terms}+\cdots=\\
=&\frac12+\frac{2}{2^2}+\frac{2^2}{2^3}+\cdots+\frac{2^k}{2^{k+1}}+\cdots\\
=&\frac12+\frac12+\frac12+\cdots=+\infty
\end{align*}
A: Yes, "$+1$" becomes negligible and it is called the Limit Comparison Test. If $$\lim_{n \to \infty} \frac{a_n}{b_n} = L \neq 0$$
then, $\sum a_n$ and $\sum b_n$ are both convergent or both divergent.
A: Use the cauchy condensation test. The series in question converges iff the series
$$
\sum \frac{2^n}{2^n+1}
$$
converges. But the latter series diverges since the terms don't go the zero (the terms go to $1$). In particular, the series in question diverges.
