In an assignment, I must prove that

if $a_n$ is a real, convergent sequence such that $\lim_{n\to\infty}a_n = a,$

then $\lim_{n\to\infty}b_n=a$ with


I think it boils down to proving that $\sum^{\infty}_{k=1}a_k$ converges iff $a_k$ is a null sequence. Then we have $\lim_{n\to\infty}b_n=0\cdot($some real number$)=0$

So, my question is:

Given $\lim_{k\to\infty}a_k=0$. How to prove, that $\sum^{\infty}_{k=1}a_k$ converges? Or is this approach incorrect?

And another question: Consider the case $a_n=\frac{1}{n}$. This is a null sequence, but the harmonic series diverges.

So we have $b_n=0\cdot\infty$. Does $\lim_{n\to\infty}b_n$ exist in this case?

  • $\begingroup$ Do you want all your sums to be ${1\over n}\sum_{k=1}^\infty a_k$? $\endgroup$ Nov 28, 2015 at 13:21
  • $\begingroup$ @DavidMitra: No, there is also $\frac{1}{\infty}$. That would be $\lim_{n\to\infty}(\frac{1}{n}\sum^{n}_{k=1}a_k)$ $\endgroup$
    – Arthur
    Nov 28, 2015 at 13:26
  • $\begingroup$ I must be missing something. What is your definition of "null sequence"? Why is $(a_n)_n$ defined by $a_n = 1/n$ one of them? $\endgroup$
    – Clement C.
    Nov 28, 2015 at 13:28
  • $\begingroup$ Oh, sorry, I meant the upper limit to be "$n$". $\endgroup$ Nov 28, 2015 at 13:30
  • 1
    $\begingroup$ Then the statement you are trying to prove is false. Convergence of $a_n$ to zero is necessary for $\sum_{n=1}^\infty a_n$ to exist, but is not sufficient -- your example of the harmonic series shows it. (Also, in terms of vocabulary, I personally associate "null sequence" to "sequence identically zero", or "eventually identically zero".) $\endgroup$
    – Clement C.
    Nov 28, 2015 at 13:31

1 Answer 1


Since $\lim\limits_{n\rightarrow\infty} a_n = a$, then $\forall\varepsilon>0, \exists n_0\in \mathbb{N}$ such that $n>n_0\Rightarrow |a_n-a|<\varepsilon/2$.

For $n>n_0$, we have \begin{align}|b_n-a|&=\left|\frac{1}{n}\left(\sum_{k=1}^{n}a_k\right)-a\right|\\ &=\left|\frac{1}{n}\left(\sum_{k=1}^{n_0}a_k+\sum_{k=n_0+1}^{n}a_k\right)-a\right|.\end{align}

There is a mean value $\alpha_n$ of $a_k$'s, $k\in\{n_0+1,...,n\}$, and since $|a_k-a|<\varepsilon/2$, then also $|\alpha_n-a|<\varepsilon/2$. Replacing $\alpha_n$ in the equation above gives \begin{align}|b_n-a|&=\left|\frac{1}{n}\left(\sum_{k=1}^{n_0}a_k+(n-n_0-1)\alpha_n\right)-a\right|\\ &=\left|\frac{1}{n}\left(\sum_{k=1}^{n_0}a_k-(n_0+1)\alpha_n\right) + \alpha_n-a\right|\\ &\leqslant\left|\frac{1}{n}\left(\sum_{k=1}^{n_0}a_k-(n_0+1)\alpha_n\right)\right| + \left|\alpha_n-a\right|\\ &<\left|\frac{1}{n}\left(\sum_{k=1}^{n_0}a_k-(n_0+1)\alpha_n\right)\right|+\frac{\varepsilon}{2}.\end{align} For $N\gg n_0$, we can assure that $$\left|\frac{1}{N}\left(\sum_{k=1}^{n_0}a_k-(n_0+1)\alpha_N\right)\right|<\frac{\varepsilon}{2},$$ because the term between the parenthesis is limited. Therefore, for $N\gg n_0$, $|b_n-a|<\varepsilon$.

Now, I don't think your approach is good. The "iff" in your sentence is not valid.

Regarding the harmonic series: If my proof is correct, then, of course, $b_k$ for the sequence $1/k$ has a limit and it is 0.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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