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The series is the sum from $n=1$ to infinity of $\frac {2n}{(8n+11)}$

I know this series diverges by the divergence test because the limit is $\frac 14$ (not zero) but how can I know what it diverges to? is it infinity?

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    $\begingroup$ If $\lim_{n\to\infty}a_n>0$, then $\sum_{n=1}^Na_n\to+\infty$ $\endgroup$ – Simply Beautiful Art Mar 24 '17 at 21:59
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    $\begingroup$ "Divergent" means that it doesn't converge, so it either goes to $\pm\infty$ or oscillates between values. $\endgroup$ – The Count Mar 24 '17 at 22:01
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    $\begingroup$ Does the series move in one direction (positive vs. negative) or does it change directions? If the series diverges, and it does not change direction, then it must go to... $\endgroup$ – abiessu Mar 24 '17 at 22:12
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    $\begingroup$ @TheCount "oscillate between values" may be a problematic intuition. for example take an enumeration $(q_n)_n$ of $\Bbb Q$, then $\sum_{n=1}^N (q_{n+1}-q_n)=q_{N+1}-q_1$ is not really oscillating (but generates a dense set) $\endgroup$ – tofurind Mar 24 '17 at 22:20
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    $\begingroup$ @TheCount Yes, I were aware of that. ;) My comment were more meant as additional information for the OP not to memorize a maybe wrong interpretation of your comment. $\endgroup$ – tofurind Mar 24 '17 at 22:44
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As you said, $\lim_{n\to\infty} \frac{2n}{8n+11} = \frac{1}{4}$, therefore the series does not converge. To answer your question we need to make two observations: the series is a sum of positive terms, and the sequence $\frac{2n+2}{8n+19}$ is an increasing one. In such a case, the series diverges to positive infinity for its value is arbitrarily large.

To see that $\frac{2n}{8n+11}$ is positive, it is sufficient to observe that n goes from 1 to infinity. On the other hand, we also see that $\frac{2n+2}{8n+19}>\frac{2n}{8n+11} \Leftrightarrow (2n+2)(8n+11)-2n(8n+19)>0 \Leftrightarrow 22>0$, thus we have what we aimed for.

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One way to investigate the convergence/divergence of a series is to use the comparison test. For this series, one can compare: $\frac{2}{19}=\frac{2n}{19n} \leq \frac{2n}{8n+11} \leq \frac{2n}{8n}=\frac{1}{4}$. Now summing:

$$\sum_{n=1}^ \infty \frac{2}{19} \leq \sum_{n=1}^ \infty \frac{2n}{8n+11} \leq \sum_{n=1}^ \infty \frac{1}{4} \Rightarrow \infty \leq \sum_{n=1}^ \infty \frac{2n}{8n+11} \leq \infty $$

Another way is (mentioned by AlexT above) to see that it is an increasing positive series: $a_n<a_{n+1}.$

Now, here is a challenge for you: Does a decreasing positive series converge: $a_n>a_{n+1}>0$?

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