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I know that this is a very simple question, but I'm struggling to find a convincing way tho show this. I'm trying to show that $\sum_\limits{n=1}^\infty \frac{1}{n+1}$ diverges. I tried using the Ratio Test and the Root Test, but both of these are inconclusive. I know that $\sum_\limits{n=1}^\infty \frac{1}{n}$ diverges, but I don't know how to show it for $n+1$.

Additionally, I would like to show that $\sum_\limits{n=1}^\infty \frac{1}{(n+1)^2}$ converges, but again, using the Ratio Test and the Root Test gives an inconclusive result. Thanks in advance.

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    $\begingroup$ You can use the Integral Test on both. $\endgroup$ – TorsionSquid Dec 4 '16 at 15:55
  • $\begingroup$ $\sum_{n=1}^{\infty}\frac{1}{n+1} = \sum_{n=2}^{\infty}\frac{1}{n} = \sum_{n=1}^{\infty}\frac{1}{n} - 1$ $\endgroup$ – Plopperzz Dec 4 '16 at 15:55
  • $\begingroup$ It is strictly decreasing, so the sum converges iff the integral. $\endgroup$ – GLay Dec 4 '16 at 15:55
  • $\begingroup$ You can use the limit comparison test, and it is possibly the best test to apply here. $\endgroup$ – ThePortakal Dec 4 '16 at 16:28
  • $\begingroup$ harmonic series $\endgroup$ – Alucard Dec 4 '16 at 19:09
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the sum $$\sum_{i=1}^n\frac{1}{n+1}$$ diverges since $$\frac{1}{n+1}\geq \frac{1}{2n}$$ for all $n$ with $$n\geq 1$$ the sum $$\sum_{i=1}^n\frac{1}{(n+1)^2}$$ converges since we have $$\frac{1}{(n+1)^2}\le \frac{1}{n^2}$$ for all $n$ with $$n\geq 1$$

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Just open the summation to figure out:

$\sum_\limits{n=1}^\infty \frac{1}{n} = 1+\sum_\limits{n=1}^\infty \frac{1}{n+1}$

$\sum_\limits{n=1}^\infty \frac{1}{n^2} = 1+\sum_\limits{n=1}^\infty \frac{1}{(n+1)^2}$

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  • To preface, the p-series test states that given

    $$ \sum_{n=1}^\infty \frac1{n^p} $$

    if $p > 1$, then the series converges. This can be proven with the integral test, which can be proven geometrically, but that is beyond the scope of this answer.

  • Also, we will use the observation that given the notation

    $$ \sum_{n=1}^\infty a_n $$

    if two series have the same end behavior (that being, if two series' $a_n$ reach the same value as $n \rightarrow \infty$), then they either both converge or both diverge since it is the infinitieth terms that matter. This logic leads to the limit comparison test, but we will simply use this logic as justification.

Divergence of $\sum\limits_{n=1}^\infty \frac1{n+1}$

Let's take a look at $\frac1{n+1}$. As $n \rightarrow \infty$, $\frac1{n+1} \rightarrow \frac1n$. This means that the series

$$ \sum_{n=1}^\infty \frac1n \tag1 \label1 $$

has the same end behavior as our original series. By the p-series test, we see that $\eqref1$ diverges. Because this series has the same end behavior as the original, one can confidently assert that the original series diverges.

Convergence of $\sum\limits_{n=1}^\infty \frac1{\left ( n+1 \right )^2}$

Let's take a look at $\frac1{\left ( n+1 \right )^2}$. As $n \rightarrow \infty$, $\frac1{\left ( n+1 \right )^2} \rightarrow \frac1{n^2}$. This means that the series

$$ \sum_{n=1}^\infty \frac1{n^2} \tag2 \label2 $$

has the same end behavior as our original series. By the p-series test, we see that $\eqref2$ converges. Because this series has the same end behavior as the original, one can confidently assert that the original series converges.

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Note that $1/(n+1)$ and $1/(n+1)^2$ are decreasing and positive. Hence, by Cauchy Condensation, $∑ 2^n \frac{1}{2^{n}+1} ≥ ∑ 1/2 = \infty$ so $∑1/(n+1)$ diverges, and $∑ 2^n \frac{1}{(2^n+1)^2} ≤ ∑ \frac{1}{2^n} < ∞ $ so $∑1/(n+1)^2$ converges.

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