$\sum\left(\frac{ n^2 + 1}{n^2 +n + 1}\right)^{n^2} $ converges or diverges?

Question

The original question is to show that $\;\sum\left(\dfrac{ n^2 + 1}{n^2 +n + 1}\right)^{n^2} $ either converges or diverges.

I know it diverges but I'm having difficulty arriving at something useful for $ a_n $.

Here's what I did: $ a_n = \left(\dfrac{ n^2 + 1}{n^2 +n + 1}\right)^{n^2} \geq\left( \dfrac{ n^2 + 1}{n^2 +2n + 1}\right)^{n^2} \geq\left( \dfrac{ n^2 - 1}{n^2 +2n + 1}\right)^{n^2} = \left(\dfrac{(n-1)(n+1)}{(n+1)^2}\right)^{n^2} = \left(\dfrac{n-1}{n+1}\right)^{n^2} = \left( 1 - \dfrac{2}{n+1}\right)^{n^2} $

Do you have idea how to continue from here? ( my idea was to arrive to some expression that goes to infinity and deduce by the comparison test that $ a_n $ diverges ). I feel like something is right infront of my eyes but I can't see it.

Edit: The series converges. However, how do you propose I should continue from where I've left?

Answer 1

That series converges. If $n>1$, you have$$\left(\frac{n^2+1}{n^2+n+1}\right)^{n^2}=\left(1-\frac n{n^2+n+1}\right)^{n^2}\leqslant\left(1-\frac1{2n}\right)^{n^2}.$$But $\lim_{n\to\infty}\left(1-\frac1{2n}\right)^n=\exp\left(-\frac12\right)$. Therefore, if you take some $c\in\left(\exp\left(-\frac12\right),1\right)$, you have $\left(1-\frac1{2n}\right)^n\leqslant c$ if $n\gg1$, and so $\left(1-\frac1{2n}\right)^{n^2}\leqslant c^n$ if $n\gg1$.

Answer 2

$n^{2} \ln (1-\frac n {n^{2}+n+1}) \leq \frac 1 2 (-n^{2})\frac n {n^{2}+n+1} $ as for $n$ sufficiently large and $\sum e^{-n} <\infty$. Hence the series is convergent.

Answer 3

You need not use any test to find that it converges, just application of standard limits will do, as follows: \begin{align} \sum_{n=1}^\infty\left(\dfrac{ n^2 + 1}{n^2 +n + 1}\right)^{n^2} &=\sum_{n=1}^\infty\left[\left(1-\dfrac{1}{n +1 + \frac1n}\right)^{n+1+\frac1n}\right]^\frac{n^2}{n+1+\frac1n}\\ &=0 \end{align} This is because the inner limit is $\frac1e<1$ because $(1-\frac{1}{x})^x\xrightarrow{x\to\infty}\frac1e$ and the outer exponent goes to infinity.

Answer 4

Let's use the root test:

$$\sqrt[\Large n]{\left( \frac{n^2+1}{n^2+n+1} \right)^{n^2}} = \left( \frac{n^2+1}{n^2+n+1} \right)^{n} = \left[\left( 1-\frac{n}{n^2+n+1} \right)^{-\frac{n^2+n+1}{n}}\right]^{\frac{-n^2}{n^2+n+1}}\stackrel{n\to\infty}{\longrightarrow} e^{-1} < 1,$$

and therefore the series converges.

We used:

1. $\left( 1-\frac{n}{n^2+n+1} \right)^{-\frac{n^2+n+1}{n}}\to e$, because $\frac{n}{n^2+n+1}\to 0,$
2. $\frac{-n^2}{n^2+n+1}\to -1.$

Answer 5

$$a_n=\left(\frac{ n^2 + 1}{n^2 +n + 1}\right)^{n^2}\implies \log(a_n)=n^2\log\left(\frac{ n^2 + 1}{n^2 +n + 1}\right)$$ For large $n$ $$\log(a_n)=-n+\frac{1}{2}+\frac{2}{3 n}-\frac{3}{4 n^2}-\frac{1}{5 n^3}+O\left(\frac{1}{n^4}\right)$$ Repeat it for $a_{n+1}$ to obtain $$\log(a_{n+1})-\log(a_n)=-1-\frac{2}{3 n^2}+\frac{13}{6 n^3}+O\left(\frac{1}{n^4}\right)$$ $$\frac{a_{n+1}}{a_{n}}=e^{\log(a_{n+1})-\log(a_n)}=\frac 1e \left(1-\frac{2}{3 n^2}+\frac{13}{6 n^3} \right)+O\left(\frac{1}{n^4}\right)$$ So, it converges.

Now, if you make $$\sum_{n=0}^\infty a_n=\sum_{n=0}^p a_n+\sum_{n=p+1}^\infty a_n$$ $$\sum_{n=p+1}^\infty a_n\sim \sum_{n=p+1}^\infty e^{\frac{1}{2}-n}=\frac{e^{\frac{1}{2}-p}}{e-1}$$

If you make $p=4$, the approximation gives $$\frac{31179700687749571657625354575093}{15170464778704564632455243642733}+\frac{1}{(e- 1) e^{7/2}}\sim 2.072864$$ while the "excat" result is $2.074636$.