Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|cite|improve this question
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

2 Answers 2

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.

share|cite|improve this answer
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 $n > 2$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.