I'm trying to prove that $\displaystyle \sum_{n=1}^{\infty}\frac{n^{10}}{1.1^n}<\infty$.

My try:

Let $n_0\in \mathbb{N}$ such that for any $n>n_0$ the condition $1.1^n>n^{12}$ holds.

Now, $$\sum_{n=1}^{\infty}\frac{n^{10}}{1.1^n}=\sum_{n=1}^{n_0}\frac{n^{10}}{1.1^n}+\sum_{n=n_0+1}^{\infty}\frac{n^{10}}{1.1^n}\le\sum_{n=1}^{n_0}\frac{n^{10}}{1.1^n}+\sum_{n=n_0+1}^{\infty}\frac{n^{10}}{n^{12}}$$Both sums in RHS converge, hence $\displaystyle \sum_{n=1}^{\infty}\frac{n^{10}}{1.1^n}$ converges.

Is my reasoning fine?

By the way, According to WolframAlpha the series converges by the ratio test, but I couldn't find proper series to check with. Please explain.

Thank you!

  • 3
    $\begingroup$ When using the ratio test you do not need to pit the series against another one - that business is for comparison test (aka majorant/minorant). You only need to study the limit of the ratio of two consecutive terms. Here $$\frac{a_{n+1}}{a_n}=\frac{(n+1)^{10}}{1.1\cdot n^{10}}.$$ Oh, and your reasoning is fine. Provided that you can justify the existence of such an $n_0$. Presumably you have, at your disposal, a general result stating that exponential growth always wins against polynomial growth. $\endgroup$ – Jyrki Lahtonen Nov 17 '15 at 10:39
  • $\begingroup$ @JyrkiLahtonen, thank you for useful answer. $\endgroup$ – Galc127 Nov 17 '15 at 10:43
  • 1
    $\begingroup$ Another way of solving this is the root test and noticing that $n^{1/n} \rightarrow 1$ as $n$ goes off to $\infty$. $\endgroup$ – MathNewbie Nov 17 '15 at 10:44
  • $\begingroup$ Actually, you have found a proper series to check with for the direct comparison test -- what you're showing here is that $a_n < n^{-2}$ for sufficiently large $n$, and you already know that $\sum n^{-2}$ converges. $\endgroup$ – Henning Makholm Nov 17 '15 at 10:56

Yes, this reasoning is fine, provided that you have a reason to know that your $n_0$ exists.

For the ratio test: The ratio between successive terms in the series is $$ \frac{(n+1)^{10}/1.1^{n+1}}{n^{10}/1.1^n} = \frac{(n+1)^{10}}{n^{10}\cdot 1.1} = \frac1{1.1}\Bigl(\frac{n+1}n\Bigr)^{10} = \frac1{1.1}\Bigl(1+\frac1n\Bigr)^{10} $$ which clearly converges toward $\frac1{1.1}$ which is strictly less than $1$. Thus, the series must converge.


I would say your argument is semi-right; splitting a series whose convergence is unknown like that is "dubious".

From the first sentence below "My Try" I infer that you know the fact that for all $a,b > 0$ we have $x^{a}/e^{bx} \to 0$ as $x \to \infty$. Using this fact we have $$ \frac{n^{10}}{(1.1)^{n}} = \frac{n^{10}}{\exp [n \log (1.1)]} \leq \frac{n^{10}}{n^{12}} = n^{-2} $$ for large $n$; the series $\sum_{n \geq 1}n^{-2}$ converging implies, by comparison test, that the series under consideration converges.

  • $\begingroup$ This kind of "dubiousness" is fine, though -- the first part if the split is a plain old finite sum and obviously has a finite value. And the partial sums of the entire series are (from a point onwards) the same as the partial sums of the infinite part of the split plus the constant finite value of the first part, so if the second part of the split converges, then so does the entire series. $\endgroup$ – Henning Makholm Nov 17 '15 at 10:46
  • $\begingroup$ @HenningMakholm Oh I agree this. Unaware of Galc127's background, I intended to use the word "dubious" along with quotation marks to make it as a gentle reminder. Doing so is fine when the doer completely knows what he is doing. :) $\endgroup$ – Megadeth Nov 17 '15 at 10:49

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