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Is the series $\sum_{n=1}^{\infty} \frac{n^2}{3^n-2^n}$ convergent?

My Approach:

$$\sum_{n=1}^{\infty} \frac{n^2}{3^n-2^n}$$

$$\lim_{n\to \infty}\left|\frac{a_{n+1}}{a_n}\right|$$

So, I did this, and I will write down the simplified form because its hard to type big fractions:

$$\lim_{n\to \infty} \frac{(3^n-2^n)(n+1)^2}{(3^{n+1}-2^{n+1})n^2}=\lim_{n\to \infty} \frac{(3^n-2^n)(n+1)^2}{(3\cdot3^n-2\cdot2^n)n^2}$$

Now all I have to do is prove that the denominator gets larger faster than the numerator. But the problem is unlike other questions both the numerator and the denominator have the same highest power. So:

$$\lim_{n\to \infty} \frac{(3^n-2^n)(n^2+\cdots)}{(3\cdot3^n-2\cdot2^n)n^2}$$

Divide by $n^2$:

$$\lim_{n\to \infty} \frac{(3^n-2^n)}{(3\cdot3^n-2\cdot2^n)}$$

Other terms in the form $\frac{a}{n}$ and $\frac{b}{n^2}$ in the numerator go to zero, so I omitted writing that in latex.

Now is this clear to show that the denominator is larger than the numerator? Because 3, and 2 seems small numbers, nothing compared to powers that we deal with in other questions.

Also I know that this converges because I confirmed it on http://www.wolframalpha.com/input/?i=%5Csum_%7Bn%3D1%7D%5E%7B%5Cinfty%7D+%5Cfrac%7Bn%5E2%7D%7B3%5En-2%5En%7D

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    $\begingroup$ $\frac{3^n-2^n}{3\cdot 3^n-2\cdot 2^n}\leq \frac{3^n-2^n}{2\cdot 3^n-2\cdot 2^n} = \frac{1}{2}$ $\endgroup$ – Alex Fish Nov 2 '15 at 1:07
  • $\begingroup$ @AlexFish Wait, OHHH! You need to turn that into an answer for me to accept it :D Just copy paste it as an answer. That was smart +1 $\endgroup$ – M.S.E Nov 2 '15 at 1:13
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$$\frac{3^n-2^n}{3\cdot 3^n-2\cdot 2^n}\leq\frac{3^n-2^n}{2\cdot 3^n-2\cdot 2^n}=\frac{1}{2}$$

:)

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  • $\begingroup$ Do you imply $\frac{3^n-2^n}{3^{n+1}-2^{n+1}}<0$ for some $n$? $\endgroup$ – Alex Fish Nov 2 '15 at 1:42
  • $\begingroup$ Ohh..I think I get you. $\mid \frac{a_{n+1}}{a_n} \mid$ should be smaller than $1$ not $\mid a_n\mid$. $\endgroup$ – Alex Fish Nov 2 '15 at 1:47
  • $\begingroup$ Which limit can you calculate directly easily? $\endgroup$ – Alex Fish Nov 2 '15 at 1:57
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You should finish dividing top and bottom by $3^n$: $$\lim_{n\to\infty}\frac{1-(2/3)^n}{3-2(2/3)^n}=\frac13<1. $$ Besides equivalents, the series can be shown to converge also by comparison as follows: $$\frac{n^2}{3^n-2^n}=\frac{n^2/2^n}{(3/2)^n-1}<\frac{n^2}{2^n}<\left(\frac{1+\varepsilon}{2}\right)^n, $$ where the rightmost inequality holds for large enough $n$, however small you pick $0<\varepsilon<1$. Since we know $\sum\left(\frac{1+\varepsilon}{2}\right)^n$ converges (it's a geometric series and $\frac{1+\varepsilon}{2}<1$), we're done.

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As often, it's much shorter with equivalents:

$3^n-2^n\sim_\infty 3^n$, hence $$\frac{n^2}{3^n-2^n}\sim_\infty\frac{n^2}{3^n}, $$ and the latter is known to converge, e. g. because it is $\;o\Bigl(\dfrac1{n^2}\Bigr)$.

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  • $\begingroup$ Really you can omit $2^n$ ? That is so cool! $\endgroup$ – M.S.E Nov 2 '15 at 1:14
  • $\begingroup$ This method of proof may be helpful for others, but I do not think it is at the right level considering the question asked. That is, I doubt that the tilde_infinity and small-o notation are familiar to the OP. $\endgroup$ – Benjamin Dickman Nov 2 '15 at 1:15
  • $\begingroup$ It easy to prove, since $2^n=o(3^n)$. $\endgroup$ – Bernard Nov 2 '15 at 1:16
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    $\begingroup$ @BenjaminDickman: I don't know. Here it is part of the curriculum during the university 1st year. $\endgroup$ – Bernard Nov 2 '15 at 1:18
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    $\begingroup$ @BenjaminDickman Yes you are right :) I am not aware of that notation or what that stands for $\endgroup$ – M.S.E Nov 2 '15 at 1:23
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Another approach - just for fun! Details left to the reader.

For big enough $n$, note that $3^n - 2^n > 2^n$ because you'll have $3^n > 2^n + 2^n = 2^{n+1}$.

Specifically, this occurs for $n \geq 2$.

From this, we conclude that for $n \geq 2$:

$$\frac{1}{3^n - 2^n} < \frac{1}{2^n}$$

And so your series (discounting the first term) has all positive terms and is bounded above by:

$$\sum_{n=2}^{\infty} \frac{n^2}{2^n}$$

Clearly $2^n \geq n^4$ for big enough $n$; this occurs for $n \geq 16$.

That latter fact means discounting a few terms, your series is bounded above by:

$$\sum_{n=16}^{\infty} \frac{n^2}{n^4} = \sum_{n=16}^{\infty} \frac{1}{n^2}$$

which you may recognize as a convergent series.

Takeaways: For a series with all positive terms, it is convergent iff it is bounded above. Moreover, removing a finite number of terms does not affect convergence; when I first learned about sequences and series, we referred to this as FFTDM (First Few Terms Don't Matter). And so the question becomes, how can you manipulate the series at hand so that it is less than something that you know to converge?

(Often one tries to use an upper bound in which the series uses $1/n^p$ for $p > 1$; some texts call this a $p$-series or some such thing.)

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