Does $\sum_0^\infty(\frac{1}{9n+1})$ converges? If yes, then to what?
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Use the comparison test. – Hakim Jun 08 '14 at 11:04
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1Hint: ${1\over 9n+1}\ge {1\over 18 n}$. – David Mitra Jun 08 '14 at 11:04
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1so it diverges correct?? – Satvik Mashkaria Jun 08 '14 at 11:06
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Yes, that's right. – David Mitra Jun 08 '14 at 11:09
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No it doesen't converge. You can see this in several ways. As David Mitra pointed out you can observe that $\frac1{9n+1}\ge\frac1{18n}$; then you'll have $$ \sum_{n=0}^{+\infty}\frac1{9n+1}\ge\sum_{n=1}^{+\infty}\frac1{18n}= \frac1{18}\sum_{n=1}^{+\infty}\frac1{n}=+\infty\;\;. $$ Another way is to observe that $\frac1{9n+1}\sim_{+\infty}\frac1{9n}$ (in fact $\lim_n\frac{1/(9n+1)}{1/9n}=1)$ hence the two series $\sum\frac{1}{9n+1}$ and $\sum\frac1{9n}$ will have the same type of behaviour, so you can conclude.
Joe
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