# Convergent or divergent $\sum_{n=1}^{\infty}\frac{1}{\sqrt{n^2+1}}\left(\frac{n}{n+1}\right)^n$?

Any suggestions? I have tried using D'Alembert's test, but on the end I get 1. I can't think of any other series with which to compare it. In my textbook the give the following solution which I don't quite understand:

$\sum_{n=1}^{\infty}\frac{1}{\sqrt{n^2+1}}(\frac{n}{n+1})^n=\sum_{n=1}^{\infty}\frac{1}{\sqrt{n^2+1}}\frac{1}{(1+\frac{1}{n})^n}\sim \sum_{n=1}^{\infty}\frac{1}{n}\frac{1}{e} \sim \sum_{n=1}^{\infty}\frac{1}{n}$ and therefore it diverges.

I don't understand the meaning of $\sim$ and the hole logic behind this answer. To me this doesn't look completly rigorous. Was here any of the convergence/divergence test implictly used?

• @Nick - One thing to remember with textbook answers is that it is not usually possible, because of space limitiations, to give complete, rigorous answers. So, you get the telegraphic version that says "essentially", i.e., asymptotically, your series is a constant multiple of a divergent series, the harmonic. – Chris Leary Aug 2 '14 at 18:42

I completely agree with you. This argument does not look very rigorous, though it does provide some sort of intuition. (Edit: Note that @sami has provided a nice rigorous interpretation of the problem)

To provide a more rigorous approach, note that the limit comparison test works nicely here.

Let \begin{align*} a_n &= \frac{1}{\sqrt{n^2+1}}\left(\frac{n}{n+1}\right)^n & b_n &= \frac{1}{n} \end{align*} Then $$\lim_{n\to\infty}\frac{a_n}{b_n}=\frac{1}{e}$$ (see if you can prove this!)

The limit comparison test then implies that either both $\sum a_n$ and $\sum b_n$ converge or both diverge. Of course $\sum b_n$ is harmonic so $\sum a_n$ diverges.

The symbol $\sim$ means asymptotically equivalent and we say that two sequences $(u_n)$ and $(v_n)$ are asymptotically equivalent and we write $u_n\sim v_n$ if

$$u_n-v_n=o(u_n)\tag1$$ which's equivalent in the case $u_n\ne0$ for $n\ge n_0$ to $$\lim_{n\to\infty}\frac{v_n}{u_n}=1$$ Now from $(1)$ we see that $\sum_n u_n$ is convergent if and only if $\sum_n v_n$ is convergent.

• Then the last relation involving "~"in the question is wrong: the ratio converges to $\dfrac{1}{e}$ instead of $1$. But I agree with your definition. I guess the textbook author's only cared about if the sum is convergent or not. – anderstood Aug 2 '14 at 17:31
• I think that $a_n~b_n$ should mean $\lim_{x\to \infty} \frac{|{a_n}|}{|{b_n}|}\in \mathbb{R}_+$ – mlainz Jan 28 '16 at 21:42
• We would have that if $a_n \sim b_n$, then they have the same (convergent/divergent) behavior. – mlainz Jan 28 '16 at 21:49
• You can also define an order relation $\lesssim$ in a similar fashion. I find it a very intuitive way to understand convergence. – mlainz Jan 28 '16 at 21:52

The $∼$ symbol means that both expresions have similar behavior in infinity,

For instance consider the series $\sum a_n$ , $\sum b_n$

Where

$a_n=\frac{1}{n}$ and $b_n=\frac{1}{n}\frac{1}{e}$

So by Limit comparison test,

$\displaystyle{\lim_{x\to \infty} \frac{a_n}{b_n}=e>0}$

As we know armonic series diverges therefore $\sum _{n=1}^{\infty }\frac{1}{n}\frac{1}{e}$ diverges.

• Also in this theory is quite practical to use the fact that: $\displaystyle{\lim_{n\to \infty}\left(1+\frac{1}{n}\right)^n:=e}$ – user115442 Aug 2 '14 at 18:23
There's one more way. Consider $$a_n = \frac{1}{\sqrt{n^2+1}} (\frac{n}{n+1} )^n = e^{\log \frac{1}{\sqrt{n^2+1}} (\frac{n}{n+1} )^n} = e^{-\frac{1}{2}\log (n^2+1) +n \log \frac{n}{n+1}}$$ Here it's useful to use the Taylor expansion for the logarithm function. Asymptotically as $x \to 0, \ \log (1+x) \sim x$ and the property of logarithm: $\log a^t = t \log a$. Therefore, $$a_n = e^{-\log n} \cdot e^{-\frac{1}{2n^2}} \cdot e^{-\frac{n}{n+1}} \geq e^{-\log n } e^{-1} e^{-1} = \frac{e^{-2}}{n} =b_n$$ which, just like you have shown, diverges because $b_n \sim \frac{1}{n}$, which is known as Harmonic series.
Squeezing if for the ultimate non-believers. Since $2\leq\left(1+\frac{1}{n}\right)^n\leq e$ and $n\leq\sqrt{n^2+1}\leq(n+1)$, $$\sum_{n=1}^{N}\frac{1}{e(n+1)}\leq\sum_{n=1}^{N}a_n \leq \sum_{n=1}^{N}\frac{1}{2n},$$ but the LHS is greater than: $$\frac{1}{e}\sum_{n=1}^{N}\log\left(1+\frac{1}{n+1}\right) = \frac{1}{e}\log\frac{N+2}{2}.$$