I am almost embarrassed to asked to this question, but after considering it for a while I realize I need some help.

In the following $a_n, b_n >0$.

So, by limit comparison test if $a_n = O(b(n))$ or, for a stricter case that I am interested in, $\frac{a_n}{b_n} \to_n 1$ both $S_1 = \sum_n a_n$ and $S_2 = \sum_n b_n$ either diverge or converge or, in other words, $S_1 = O(S_2)$ or $\frac{S_1}{S_2} \to_n O(1)$.

How about a stricter condition: does this convergence imply that $\frac{S_1}{S_2} \to_n 1$ or $S_1 \sim S_2$?. For example, $S_1 \sim n \log n + O(1)$ and $S_2 \sim n \log n + n + \frac{1}{n}$.

I think the answer is not in general, I tried to compare $\sum_k \log (1+\frac{k}{n})$ and $\sum_k \frac{k}{n}$ (they are $\sim$ if Taylor series is used). And I got the first constant to be $\log \frac{4}{e}$, and the second, of course, $\frac{1}{2}$.

Any suggestions?


After reading the solution, I went through my calculation again and what I get is $$ \sum_{k=1}^{n} \log (1+\frac{k}{n}) = \log (2n)! - \log n! - n \log n $$ Either using Stirling's approximation or $\sum_{k=1}^{n} \log k \sim \int_{1}^{n} \log x dx$ I get the front term equal to $\log \frac{4}{e}$, which is different to $\frac{1}{2}$ from $\sum_{k=1}^{n} \frac{k}{n}$. Where is the mistake?

  • $\begingroup$ Are you talking about $S_N^{(1)} = \sum_{n=1}^N a_n$, and an analogously defined $S_N^{(2)}$? Or are you talking about $\sum_{n=1}^\infty a_n$? In the second case, asymptotics dont really make sense... $\endgroup$
    – PhoemueX
    Jun 30 '16 at 20:23
  • $\begingroup$ The former. Whether the series converge or not is of little interest. I am interested in the constant of the largest term $\endgroup$
    – Alex
    Jun 30 '16 at 20:30

This was the answer before OP had positivity conditions for $a_n$, $b_n$.

Recall that the limit comparison test usually applied when the terms are positive. So, by making some negative terms, we might easily get a counterexample. Consider this,

$$ a_n = (-1)^n + \frac1n $$ and $$ b_n= (-1)^n $$

Then $a_n/b_n = 1+(-1)^n / n \rightarrow 1$. However,

$$ \sum_{n\leq N} a_n = \frac{1-(-1)^N}2 + \sum_{n\leq N} \frac1n = \log N + O(1)$$ and $$ \sum_{n\leq N} b_n = \frac{1-(-1)^N}2 = O(1).$$

Answer after the positivity conditions.

Let $S_{1,N}=\sum_{n\leq N} a_n$ and $S_{2,N}=\sum_{n\leq N} b_n$.

If $a_n = O(b_n)$, then we have for some positive constant $C$, $$ a_n \leq C b_n.$$ Thus, $$ \sum_{n\leq N} a_n \leq \sum_{n\leq N} C b_n, $$ which is $S_{1,N}=O(S_{2,N})$.

Similarly if $a_n\sim b_n$, then by limit comparison test, $\sum a_n$ and $\sum b_n$ both converge, or both diverge.

Since the case both converge is not of our interest: (both being $O(1)$.)

We consider the case both diverge.

For any $\varepsilon>0$, there is $N_0$ such that if $n\geq N_0$, then $$ 1-\varepsilon \leq \frac {a_n}{b_n} \leq 1+\varepsilon.$$

Let $N> N_0$. We have by the above inequalities, $$ (1-\varepsilon)\sum_{N_0< n\leq N} b_n \leq \sum_{N_0< n\leq N} a_n \leq (1+\varepsilon)\sum_{N_0< n \leq N} b_n.$$

Then we have $$ (1-\varepsilon)(S_{2,N}-S_{2,N_0}) \leq S_{1,N}-S_{1,N_0}\leq (1+\varepsilon)(S_{2,N}-S_{2,N_0}). $$ Dividing everything by $S_{2,N}$, we have $$ (1-\varepsilon)\left( 1-\frac{ S_{2, N_0} }{S_{2,N}} \right)\leq \frac{S_{1,N}-S_{1,N_0}}{S_{2,N}}\leq (1+\varepsilon)\left( 1-\frac{ S_{2, N_0} }{S_{2,N}} \right) $$ Letting $N\rightarrow\infty$, we have $$ 1-\varepsilon \leq \liminf_{N\rightarrow\infty} \frac{S_{1,N}}{S_{2,N}} \leq \limsup_{N\rightarrow\infty} \frac{S_{1,N}}{S_{2,N}} \leq 1+ \varepsilon. $$ Since $\varepsilon>0$ is arbitrary, we obtain that $$ \lim_{N\rightarrow\infty} \frac{S_{1,N}}{S_{2,N}} = 1. $$ Therefore, in this case, $S_{1,N} \sim S_{2,N}$.

  • $\begingroup$ What you have written is not exactly an answer to my question-it is a counterexample that shows that LCT does not work for alternating series, but thanks anyway $\endgroup$
    – Alex
    Jun 30 '16 at 20:37
  • $\begingroup$ My example shows that the partial sums do not satisfy $S_{1,N} = O(S_{2,N})$. Since you did not specify that terms are positive, I think this answers your question. $\endgroup$ Jun 30 '16 at 20:41
  • $\begingroup$ OK I agree (hence upvoted), and I fixed the question as well $\endgroup$
    – Alex
    Jun 30 '16 at 20:57
  • $\begingroup$ I edited my answer after your edit. $\endgroup$ Jun 30 '16 at 21:38
  • $\begingroup$ Thanks, this make sense. Could you pls have a look at the edit I added to my question. Is there a mistake in my logic? $\endgroup$
    – Alex
    Jul 1 '16 at 9:22

I don't think it holds.

Take $ a_{n} = \frac{1}{n!} + \frac{1}{(n!)^{2}} $ and $ b_{n} = \frac{1}{n!} $. Therefore, $$ \lim_{n\to\infty} \frac{a_{n}}{b_{n}} = 1 $$ and

$$ S_{a}(N)=\sum_{n=0}^{N}a_{n} \Rightarrow \lim_{N\to\infty}S_{a}(N) = e+\alpha $$ $$ S_{b}(N)=\sum_{n=0}^{N} b_{n} \Rightarrow \lim_{N\to\infty}S_{a}(N) = e $$

However, $$ \lim_{n\to\infty} \frac{S_{a}(n)} {S_{b}(n)} = 1+\frac{\alpha}{e} $$


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