# If $a_n$ is a null sequence, does $\sum^{\infty}_{n=1}a_n$ converge?

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

$b_n=\frac{1}{n}\sum^{n}_{k=1}a_k$.

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?

• Do you want all your sums to be ${1\over n}\sum_{k=1}^\infty a_k$? – David Mitra Nov 28 '15 at 13:21
• @DavidMitra: No, there is also $\frac{1}{\infty}$. That would be $\lim_{n\to\infty}(\frac{1}{n}\sum^{n}_{k=1}a_k)$ – Arthur Nov 28 '15 at 13:26
• 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? – Clement C. Nov 28 '15 at 13:28
• Oh, sorry, I meant the upper limit to be "$n$". – David Mitra Nov 28 '15 at 13:30
• 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".) – Clement C. Nov 28 '15 at 13:31

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 $N\gg n_0$, $|b_n-a|<\varepsilon$.
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.