# Thread: if {a(n)} converges to a, (1/n)(sum k=1, n of {a(k)}) converges to a

1. ## if {a(n)} converges to a, (1/n)(sum k=1, n of {a(k)}) converges to a

i have to prove the following,

if {a(n)} converges to a, (1/n)(sum k=1, n of {a(k)}) converges to a
can you help me? :/

2. ## Re: if {a(n)} converges to a, (1/n)(sum k=1, n of {a(k)}) converges to a

Originally Posted by nappysnake
i have to prove the following,
if {a(n)} converges to a, (1/n)(sum k=1, n of {a(k)}) converges to a
There are two cases.
Case I, Suppose that $a=0$. Notation ${M_n} = \sum\limits_{k = 1}^n {\frac{{{a_k}}}{n}}$ and ${S_n} = \sum\limits_{k = 1}^n {{a_k}}$.

Then $\left| {{M_n}} \right| = \left| {\frac{{{S_n}}}{n}} \right| \leqslant \left| {\frac{{{S_n} - {S_K} + {S_K}}}{n}} \right| \leqslant \frac{1}{n}\sum\limits_{k = K + 1}^n {\left| {{a_k}} \right|} + \frac{{\left| {{S_K}} \right|}}{n}$
Now I leave the details to you.

Case II Suppose $a\ne 0$.
Then if $b_n=a-a_n$ then note that $(b_n)\to 0.$

Fall back to case I.