A Beautiful Problems -

I hope [] does not mean we are taking integral part of x

No Sir , of course not .

Ok.

Since \lim_{n \rightarrow \infty} a_n = x, given any infintesimal ε there exists a

natural number N such |a_n-x|<\epsilon for all n>N

i.e. x-\epsilon<a_n<x+\epsilon for all n>N

So, we have\frac{a_1+a_2+...+a_n}{n} = \frac{a_1+a_2+...+a_N}{n} + \frac{a_{N+1}+a_{N+2}+...+a_n}{n}

< \frac{k}{n}+\frac{(n-N)(x+\epsilon)}{n}

With a similar working we obtain the lower bound \frac{k}{n}+\frac{(n-N)(x-\epsilon)}{n}

Thus the limit of the expression is bounded by x- \epsilon,x+ \epsilon for any arbitrary ε

In other words the limit of the expression is also x.

Time to close another " flop - show " thread by posting my " useless " solution .

Let , " c _n , d _n ( n ≥ 1 ) , " be two sequenses of real numbers , and

c _n = a ₁ + a ₂ + ....................a _n

d _n = n

Obviously , lim [ n → ∞ ] ( c _n - c _{n - 1}d _n - d_{n - 1} ) = x

By Cesaro - Stolz theorem , lim [ n → ∞ ] ( c _nd _n ) = x

which is precisely what we had to prove .

lim_{n→∞} a₁ + a₂+....a_nn = lim_{n→∞}a₁ + a₂+....a_n-1n-1

let S = a₁+a₂...a_n-1

so lim_{n→∞} S + a_nn=Sn-1

so lim_{n→∞} S = (n-1)a_n

so lim_{n→∞} S+a_nn = lim_{n→∞} a_n = x

"Time to close another " flop - show " thread by posting my " useless " solution . "

you are made to say that simply because things like "Cesaro - Stolz theorem" and all arent in jee syllabus. Btw its good that you post good questions, but problem is that they arent jee syllabus , and hence most people here wont touch your problem as most of them( I for one) dont know even the C of "Cesaro - Stolz theorem". Now someone will pop up and say that some of these problems can be solved by using methods in jee syllabus, i agree, but some problems are of the type that jee never asks .

Yeah , I guess so . Thanks anyway for informing me of my " wrongdoings " - :)

no these are not your wrong doings, but actual fact is that you are way above our level ( or most of us' ) [6]