Find if the series converges or diverges, $a_n=\sum_{n=1}^\infty\frac{1}{1+\ln (n)}$. Comparing it with another series $b_n=\frac{1}{\ln(n)}$. Dividing both the series and taking limits, we get $\lim_{n\to\infty}\frac{\ln(n)}{1+\ln(n)}$. Since it is the $\infty/\infty$ form, applying H'opitals rule, we get, $\lim_{n\to\infty}\frac{1/n}{1/n}=1$. Now, $\lim_{n\to\infty}\frac{1}{ln(n)}=0, \Rightarrow b_n$ converges $\Rightarrow a_n$ converges. But the answer is, Comparing it with $\frac{1}{n}$(divergent harmonic series) we get,$\lim_{n\to\infty}\frac{n}{1+\ln(n)}=\lim_{n\to\infty}\frac{1}{1/n}=\lim_{n\to\infty}n=\infty \Rightarrow a_n $ diverges, what is wrong with my comparison?

  • $\begingroup$ There's something confusing here: you define $a_n$ as a quantity that doesn't depend on $n$. You probably meant that $a_n$ is the $n^{th}$ term of the series. And, for the limit comparison test (which I assume you're using), you need to know whether $\sum b_n$ converges or not. $\endgroup$ – Patrick Apr 4 '12 at 17:37
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    $\begingroup$ In fact $\sum_{2}^\infty \frac{1}{\log n}$ diverges. You seem to be saying that if $\lim_{n\to \infty}a_n=0$, then $\sum_{n=1}^\infty a_n$ converges. This is not rue. $\endgroup$ – André Nicolas Apr 4 '12 at 17:56
  • $\begingroup$ Proof that $\sum_2^\infty$ $1\over\log_bn$ diverges: $\log_bn < n$(at least for $n >$ some value) so $1\over\log_bn $$>$$1\over{n}$ and $\sum$$1\over{n}$ is divergent so $\sum$$1\over\log_bn$ is divergent $\endgroup$ – Justin Apr 26 '13 at 23:29

Your mistake is when you say that: $$\lim_{x \to \infty} b_n = 0$$ implies that $\sum b_n$ is convergent. That doesn't hold in general.

Note further that $n > \ln(n)$ so $\frac{1}{n} < \frac{1}{\ln(n)}$. So by comparison $\sum \frac{1}{\ln(n)}$ is divergent.

So as mentioned in the comment above, your comparison is fine, the conclusion is just that the series is divergent.


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