Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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. –  Patrick Apr 4 '12 at 17:37
1  
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. –  André Nicolas Apr 4 '12 at 17:56
    
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 –  Quincunx Apr 26 '13 at 23:29
add comment

1 Answer

up vote 3 down vote accepted

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.