# Checking the convergence of series

Here is the given series 3/(9n+1), decide whether it converges or diverges. I used the ratio test only to end up with the ratio=1. I know this is harmonic series but it is smaller than 1/n, therefore i cannot conclude it diverges. Please help!!

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The ratio and root tests are very crude and won't work here. Have you tried comparison oar limit comparison?

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with 1/n you mean? –  d13 Oct 28 '12 at 14:53
because then using limit comparison test with 1/n i found it to be divergent. is that right? –  d13 Oct 28 '12 at 14:55
Yes. That is what I had hoped you would do. I didn't want to "spoil" the problem for you. –  ncmathsadist Oct 28 '12 at 22:19
ahh i see, thanks a lot!! –  d13 Oct 29 '12 at 11:21

$$\frac{3}{9n+1} \ge \frac{1}{9n+1} \ge \frac{1}{9n+9} \ge \frac{1}{9(n+1)} \ge \frac{1}{9} \frac{1}{n+1}$$

At which point you should be able to figure that out...

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are u trying to show that this series is greater than the divergent series therefore it is divergent? –  d13 Oct 28 '12 at 15:01

We have $$\frac{3}{9n+1}\geq \frac{3}{9n+9}=\frac{1}{3(n+1)}.$$ Since $\displaystyle\sum_{n=1}^{\infty}\frac{1}{3(n+1)}$ is disvergent, $\displaystyle\sum_{n=1}^{\infty}\frac{3}{9n+1}$.

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thankyou very much for clearing my doubts. –  d13 Oct 28 '12 at 15:09
@d13: You are welcome. –  blindman Oct 28 '12 at 15:18

Hint

$$\dfrac{3}{9 n+1} \geqslant \dfrac{3}{12n}=\dfrac{1}{4}\cdot \dfrac{1}{n}$$

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