Given $\sum _{n=1}^{\infty} \left(\frac{n^{2} + 1}{n^{2} +n +1}\right)^{n^2}$, prove that it converges.

I tried to use the Ratio test. I got a terrible algebraic expression: $$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \left(\frac{n^4 +2n^3 +3n^2 + 2n +2}{n^4 + 4n^3 + 7n^2 + 6n +3}\right)\cdot \left(\frac{n^2 + 2n + 2}{n^2 + 3n + 3}\right)^{2n +1}$$

Now problem is I'm not allowed to use L'Hoptial and I don't really know what to do with this disaster.

  • $\begingroup$ Tip: Note that the summands are equal to $(1-\frac{n}{n^2+n+1})^{n^2}$. $\endgroup$ – Serg Dec 8 '18 at 15:44

Try the root criterion.

Prove that $$\sqrt[n]{\left(\frac{n^2+1}{n^2+n+1}\right)^{n^2}}=\left(1-\frac{n}{n^2+n+1}\right)^n$$ and that this tends to $L<1$ as $n\to \infty$. Then, the criterion implies that the sum converges.

To find the limit, remember that if $a_n\to 0$ then $$(1+a_n)^{\frac1{a_n}}\to e.$$

So $$\left(1-\frac{n}{n^2+n+1}\right)^n=\left(1-\frac{n}{n^2+n+1}\right)^{\left(-\frac{n^2+n+1}{n}\right)\cdot \left(-\frac{n}{n^2+n+1}\right)\cdot n}=$$ $$=\left[\left(1-\frac{n}{n^2+n+1}\right)^{-\frac{n^2+n+1}{n}}\right]^{ \left(-\frac{n}{n^2+n+1}\right)\cdot n}.$$

But the bracketed expression is exactly of the form $$(1+a_n)^{\frac1{a_n}},\quad a_n\to 0,$$ so it tends to $e$. And the expression in the exponent clearly goes to $-1$, so the limit is $e^{-1}<1$.

  • $\begingroup$ yes, I tried this version too. couldn't prove that this goes to 0, or even just smaller then 1. $\endgroup$ – Jneven Dec 8 '18 at 16:25
  • $\begingroup$ Check the update. $\endgroup$ – Alejandro Nasif Salum Dec 8 '18 at 19:42

Another approach using the Root test.

Let $L = \lim_{n\to\infty} \sqrt[n]{\left(\frac{n^2+1}{n^2+n+1}\right)^{n^2}} $

$$ \begin{align} L &= \lim_{n\to\infty} \exp \log \sqrt[n]{\left(\frac{n^2+1}{n^2+n+1}\right)^{n^2}} \\ &= \exp \lim_{n\to\infty} \frac{\log \left(\left( \frac{n^2+1}{n^2+n+1} \right)^{n^2}\right)}{n} \\ &= \exp \lim_{n\to\infty} \frac{n^2 \log \frac{n^2+1}{n^2+n+1}}{n} \\ &= \exp \lim_{n\to\infty} n \log \frac{n^2+1}{n^2+n+1} \tag{1} \\ &= \exp(-1) \\ \end{align}$$

and since $L = \exp(-1) < 1$ the series is absolutely convergent by the root test.

* (1) can be shown by the fact that the Laurent expansion for $n\log\frac{n^2+1}{n^2+n+1}$ at $n=\infty$ is $-1 + \textrm{O}(\frac{1}{n})$.

  • $\begingroup$ how did you come up to the the term exp(-1) from $n \ log {\frac{n^2 +1}{n^2 + n +1}}$? that's exactly the problem $\endgroup$ – Jneven Dec 8 '18 at 16:35
  • $\begingroup$ @Jneven see my edit $\endgroup$ – Dando18 Dec 8 '18 at 16:50
  • $\begingroup$ Laurent expansion is much higher level of mathematics then what i'm famiilair with, at this moment. this is of course a solid solution - but as I don't get it, I can't solve my problem using it... $\endgroup$ – Jneven Dec 9 '18 at 11:08

By Taylor's expansion we have

$$\left(\frac{n^{2} + 1}{n^{2} +n +1}\right)^{n^2}=\left(1-\frac{n}{n^{2} +n +1}\right)^{n^2}=e^{n^2 \log\left(1-\frac{n}{n^{2} +n +1}\right)}=e^{n^2 \left(\frac{-n}{n^{2} +n +1}+O(1/n^2)\right)}\sim \frac c{e^n}$$

and then refer to limit comparison test.

  • 1
    $\begingroup$ that's perfect! incredibly aesthetic! $\endgroup$ – Jneven Dec 8 '18 at 16:27
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    $\begingroup$ @Jneven I'm glad you appreciate that so much! Thanks :) $\endgroup$ – gimusi Dec 8 '18 at 16:28

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