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Is this series convergent?

$$\sum_{k=0}^{\infty}\frac{k^{3}}{3^{k}+1}$$

Used ratio test and it got very long as you can see, is it correct if I did it like that in the exam, and is there an easier way than ratio test for this? Please do tell me both these questions! :)

$$\lim_{k\rightarrow \infty}\left|\frac{\frac{(k+1)^{3}}{3^{k+1}+1}}{\frac{k^{3}}{3^{k}+1}}\right|=\lim_{k\rightarrow \infty}\left|\frac{(k+1)^{3} \cdot (3^{k}+1)}{(3^{k+1}+1) \cdot k^{3}}\right|= \lim_{k\rightarrow \infty}\left|\frac{(k+1)^{3}}{k^{3}} \cdot \left (\frac{3^{k}+1}{3^{k+1}+1}\right)\right|=$$

$$=\lim_{k\rightarrow \infty}\left|\left(\frac{k+1}{k}\right)^{3}\cdot \left(\frac{3^{k}+1}{3^{k+1}+1}\right)\right|= \lim_{k\rightarrow \infty}\left|\left(1+\frac{1}{k}\right)^{3} \cdot \left(\frac{3^{k}+1}{3^{k+1}+1}\right)\right|=$$

$$=(1+0)\cdot(0)=0<1$$

Thus the series is convergent.

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    $\begingroup$ The second factor converges to $\frac 13$, not $0$, which should do the trick anyway. $\endgroup$ – Stefan4024 Oct 12 '16 at 18:14
  • $\begingroup$ you should bound the series by $\sum_{k=0}^{\infty} \frac{k^3}{3^{k}}$ and do whatever test you want after. $\endgroup$ – felasfa Oct 12 '16 at 18:15
  • $\begingroup$ $\lim_{k \to \infty} \frac{3^k + 1}{3^{k+1} + 1} = \lim_{k\to \infty} \frac{1 + 3^{-k}}{3 + 3^{-k}} = \frac{1}{3}$. $\endgroup$ – dannum Oct 12 '16 at 18:15
  • $\begingroup$ To simplify the calculations somewhat, you could first note that $k^3/(3^k+1) \lt k^3/3^k$. $\endgroup$ – dxiv Oct 12 '16 at 18:16
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It is correct, and you can apply the ratio test to the series $$\sum_{k=0}^\infty\frac{k^3}{3^k}$$ which is bigger and converges, too. The proof is a bit shorter.

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let $u_k$ be the general term.

we have $\lim_{k\to+\infty}k^2u_k=0$

( cause the exponential is faster than the polynomial).

thus

for enough large $k$,

$k^2u_k\leq1$.

so

$0\leq u_k \leq \frac{1}{k^2}$

which proves that the series

$\sum u_k $ converges.

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An easier way to prove it was the root test, no longer than a line, but it converges anway.

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depending on the geometric series, we get $${\displaystyle {\frac {x(x^2+4x+1)}{(1-x)^{4}}}=\sum _{k=1}^{\infty }k^3x^{k}}$$ at $x=\frac{1}{3}$ $$\sum _{k=1}^{\infty }k^3x^{k}=\sum _{k=1}^{\infty }\frac{k^3}{3^k}>\sum_{k=0}^{\infty}\frac{k^{3}}{3^{k}+1}$$ so the series is convergent

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