Convergence of $\displaystyle\sum\frac{n^5}{2^n}$

[alexmahone]

Does the following series converge?

$\displaystyle\sum\frac{n^5}{2^n}$

 

[fawaz1]

Well what tests did you try?the ratio test and the root test both give you what you need. Try them. If you need more hints let us know.Mohammad

 

[alexmahone]

Well what tests did you try?the ratio test and the root test both give you what you need. Try them. If you need more hints let us know.Mohammad

$\displaystyle\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to\infty}\frac{(n+1)^5}{2n^5}=\frac{1}{2}$

So, $\displaystyle\sum\frac{n^5}{2^n}$ converges.

 

[Krizalid1]

$a_n=\dfrac{n^k}{2^n},$ so by the root test the series always converges for all $k.$

 

[blue_raver22]

Based on the ratio test, the given series is convergent. The ratio test states that if $\displaystyle\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|<1$, then the series $\displaystyle\sum a_n$ is convergent.

Applying this test to the given series, we have:

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(n+1)^5}{2^{n+1}}}{\frac{n^5}{2^n}}\right|=\lim_{n\to\infty}\left|\frac{(n+1)^5}{n^5}\cdot\frac{2^n}{2^{n+1}}\right|=\lim_{n\to\infty}\left|\frac{(n+1)^5}{n^5}\cdot\frac{1}{2}\right|=\frac{1}{2}<1$

Therefore, by the ratio test, the series $\displaystyle\sum\frac{n^5}{2^n}$ is convergent. This means that as we add more terms to the series, the overall sum will approach a finite value. In other words, the terms in the series decrease at a fast enough rate to ensure convergence.