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Let $a_n$ be a decreasing sequence of nonnegative real numbers.

Prove that if $\sum a_n$ converges, then $na_n \to 0$.

Hint: use that $n\, a_{2n} \le a_{n+1}+\cdots + a_{2n}$

I couldn't prove this using the given hint, could someone give me a few tips?

I also have two more questions:

  • Suppose I have $2na_{2n},(2n+1)a_{2n+1}\to 0$ is that enough to say that $na_n\to 0$?

  • Is there any easy way to show $n \, a_{2n}\le a_{n+1}+\cdots + a_{2n}$, there's probably a simple inductive proof I couldn't get.

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marked as duplicate by vadim123, Did real-analysis Feb 10 '16 at 19:46

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    $\begingroup$ Related : math.stackexchange.com/questions/369669/… $\endgroup$ – Watson Feb 10 '16 at 18:47
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    $\begingroup$ How do you get $\leq 2a_{n+1}$? It's not possible for $2(a_{n+1}+\text{ positive values})\leq 2a_{n+1}$. $\endgroup$ – Thomas Andrews Feb 10 '16 at 18:48
  • $\begingroup$ @ThomasAndrews Whoops, I meant $2na_{n+1}$, but in that case I can say nothing... $\endgroup$ – YoTengoUnLCD Feb 10 '16 at 18:50
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Let $(R_N)_N$ be the sequence of remainders of your series, namely $$\forall N\in\mathbb{N},\ R_N=\sum_{n=N+1}^{+\infty}a_n.$$ Since your series converges, the sequence $(R_N)_N$ is well defined and $$\lim_{N\to+\infty}R_N=0.$$ Now, since your $a_n$'s are non-negative and the sequence is non-increasing, $$\forall n\in\mathbb{N},\ na_{2n}\leq a_{n+1}+\cdots+a_{2n}\leq R_n.$$ By the Squeeze Theorem, $$\lim_{n\to+\infty}na_{2n}=0.$$

For the odd subsequence: write, for $n\in\mathbb{N}$, $$(2n+1)a_{2n+1}=2na_{2n+1}+a_{2n+1}\leq2na_{2n}+a_{2n+1}$$ and conclude (by the Squeeze Theorem again) that $$\lim_{n\to+\infty}(2n+1)a_{2n+1}=0.$$

Finally, you have a sequence $(na_n)_{n\in\mathbb{N}}$ such that the odd and even subsequences have a nil limit: you can conclude that the sequence $(na_n)_{n\in\mathbb{N}}$ has a nil limit.


Regarding your questions:

  • Suppose I have $2na_{2n}\to0$ and $(2n+1)a_{2n+1}\to0$; is that enough to say that $na_n\to0$?

Yes: a sequence has a limit $\ell$ if and only if its odd and even subsequences have the same limit, equal to $\ell$. Apply this result to the sequence $(na_n)_n$.

  • Is there any easy way to show $na_{2n}\leq a_{n+1}+\cdots+a_{2n}$, there's probably a simple inductive proof I couldn't get.

You don't need induction here. Since the sequence $(a_n)_n$ is non-decreasing, $$\forall n\in\mathbb{N}^*,\ a_{n+1}\geq a_{n+2}\geq\cdots\geq a_{2n},$$ hence $$a_{n+1}+\cdots+a_{2n}=\sum_{k=n+1}^{2n}a_k\geq\sum_{k=n+1}^{2n}a_{2n}=na_{2n}.$$

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  • $\begingroup$ Could you explain how to get $R_n\to 0$? $\endgroup$ – YoTengoUnLCD Feb 10 '16 at 19:40
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    $\begingroup$ @YoTengoUnLCD Sure: Call $(S_N)_N$ the sequence of partial sums, namely $$S_N=\sum_{n=0}^{N}a_n,$$ and call $S$ the sum of the series, namely $$S=\lim_{N\to+\infty} S_N.$$ We clearly have: $$\forall N\in\mathbb{N}, S=S_N+R_N$$ or, equivalently, $R_N=S-S_N$. Hence $$\lim_{N\to+\infty}R_N=S-S=0.$$ $\endgroup$ – gniourf_gniourf Feb 10 '16 at 19:43
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    $\begingroup$ Oh, of course! Thanks a lot gniourf. $\endgroup$ – YoTengoUnLCD Feb 10 '16 at 19:46

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