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!!
Checking the convergence of series
1
$\begingroup$
sequences-and-series
4 Answers
1
Hint
$$\dfrac{3}{9 n+1} \geqslant \dfrac{3}{12n}=\dfrac{1}{4}\cdot \dfrac{1}{n}$$
3
$$ \frac{3}{9n+1} \ge \frac{1}{9n+1} \ge \frac{1}{9n+9} \ge \frac{1}{9(n+1)} \ge \frac{1}{9} \frac{1}{n+1} $$
At which point you should be able to figure that out...
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0are u trying to show that this series is greater than the divergent series therefore it is divergent? – 2012-10-28
2
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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1@d13: You are welcome. – 2012-10-28
1
The ratio and root tests are very crude and won't work here. Have you tried comparison oar limit comparison?
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0with 1/n you mean? – 2012-10-28
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0because then using limit comparison test with 1/n i found it to be divergent. is that right? – 2012-10-28
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1Yes. That is what I had hoped you would do. I didn't want to "spoil" the problem for you. – 2012-10-28
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0ahh i see, thanks a lot!! – 2012-10-29