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I've got a problem here:

$$\sum_{n=1}^{\infty} \frac{5^n}{n(3^{n+1})}$$

I've used the ratio test and essentially did this:

$$\sum_{n=1}^{\infty} \left( \frac{5^{n + 1}}{n (3^{n+1+1})} / \frac{5^n}{n(3^{n+1})}\right) = \frac{5^n\,5}{9(n+1)3^n} \cdot \frac{n\,3^n}{5^n}$$

With a bunch of cancellations we get $\dfrac{5n}{9n+9}$, which means it converges, as $\frac{5}{9} < 1$. But the answer says it diverges! I even tried the root test and got the same result. Where am I going wrong?

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The cancellations are not done correctly. Where you have $9$ one should have $3$. – André Nicolas Jul 24 '14 at 18:37
The limit of the inside terms do not go to zero. – Joel Jul 24 '14 at 18:38
That summation sign before the nasty fraction shouldn't be there. It's difficult to tell if this is where you are going wrong, because your algebra is wrong in any case! – TonyK Jul 24 '14 at 18:42
up vote 3 down vote accepted

$$\begin{align} a_n &= \frac{5^n}{n(3^{n+1})} \\[2em] \frac{a_{n+1}}{a_n} &= \frac{5^{n+1}}{(n+1)(3^{n+2})} / \frac{5^n}{n(3^{n+1})} \\[0.5em] &= \frac{5^{n+1} \cdot n \cdot 3^{n+1}}{5^n \cdot (n+1) \cdot 3^{n+2}} \\[0.5em] &= \frac{5}{3} \left( \frac{n}{n+1} \right) \\[0.5em] &= \frac{5}{3} \left( 1 - \frac{1}{n+1} \right) \\[0.5em] &\to \frac{5}{3}>1 \end{align}$$

Therefore, from the ratio test,we conclude that the series diverges.

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Oh my dear lord I missed a 3. I've been struggling with this for awhile. Thanks! – Pejman Poh Jul 24 '14 at 18:49

Hint: $\dfrac{5^n}{n3^{n+1}} > \dfrac{1}{n}$ for $n > 2$

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