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I have used the ratio test for series.

Let $a_n = \frac{n^4}{4^n}$

Therefore $\lim_{n\to\infty}| \frac{a_{n+1}}{a_n} |= \lim_{n\to\infty} \frac{(n+1)^4}{4^{(n+1)}} * \frac{4^n}{n^4}$

= $\lim_{n\to\infty} \frac{(n+1)^4}{4n^4} = |\frac{1}{4} | $

Since the ratio is less than 1, the series converges.

Is this correct?

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Indeed it is (there is a typo on the second line, you have $a^n$ when you should have $a_n$). – David Mitra Sep 27 '12 at 3:52
Correct. Why the absolute value around $\frac{1}{4}$? Maybe they want you to prove that the limit of $\frac{(n+1)^4}{n^4}$ is $1$. Maybe not. – André Nicolas Sep 27 '12 at 3:54
Interesting. The question only asked if it converges. Thanks! – JackReacher Sep 27 '12 at 5:14
up vote 2 down vote accepted

Yes, what you have done is correct. Alternatively, we can also use root test as follows: $$\lim_{n\to\infty}|a_n|^{\frac{1}{n}}=\lim_{n\to\infty}\left(\frac{n^4}{4^n}\right)^\frac{1}{n}=\lim_{n\to\infty}\frac{(n^{\frac{1}{n}})^4}{4}=\frac{1}{4}$$ since $\lim_{n\to\infty}n^{\frac{1}{n}}=1$. Since $\lim_{n\to\infty}|a_n|^{\frac{1}{n}}=\frac{1}{4}<1$, by root test, the series $\sum a_n$ converges.

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One way to find out (yes, it is cheating ...) is to fire up sympy (I do use the interface isympy, which uses ipython) and type:

In [1]: summation( n**4/4**n, (n,0,oo))

In [2]: 

I leave the easy proof for you ... But think about this generalization: $$ \sum_{i=0}^{\infty} \frac{ P(i) }{4^i} $$ should converge for any polynomial $P$, and thinking about why should help you!

An example:

In [2]: summation( (n**4 + n**8)/4**n, (n,0,oo))
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First time I have heard of sympy. Seems like a great resource! Thanks for your input! – JackReacher Sep 27 '12 at 8:22

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