Is there any theorem like, that the limit of the average value of an infinite series takes the same value as the original sequence?
Let lim n->infinity (an) = a be given(i.e. converges)
Then the sequence bn is defined as follows,
lim n->infinity (a1+a2+.......+an)/n
We need to comment on the convergence/divergence of bn.
So, lim n->infinity (a1+a2+.......+an)/n = lim n->infinity (an/n)= [lim n->infinity (an)]/[lim n->infinity (n)] = a/infinity ->0
==>bn converges
Or is there anything such as that the limit of the average value of an infinite series takes the same value as the original sequence?
Yes, there is a theorem that states that the limit of the average value of an infinite series takes the same value as the original sequence. Let's break down the proof for this theorem:
Suppose we have a sequence (an) that converges to a limit "a" as n approaches infinity, i.e., lim n->infinity (an) = a.
Now, let's define another sequence (bn) as the average of the first n terms of the original sequence, i.e., bn = (a1 + a2 + ... + an) / n.
To determine the convergence or divergence of sequence (bn), we can take the limit as n approaches infinity of bn.
lim n->infinity (a1 + a2 + ... + an)/n = lim n->infinity (an/n).
Using the limit properties, we can split the limit of a sum into the sum of limits:
lim n->infinity (an/n) = lim n->infinity (a1/n + a2/n + ... + an/n).
Then, we can take the limit of each term separately:
lim n->infinity (a1/n + a2/n + ... + an/n) = lim n->infinity (a1/n) + lim n->infinity (a2/n) + ... + lim n->infinity (an/n).
Since the limit of the constant term divided by n as n approaches infinity is 0, we can rewrite the expression as:
lim n->infinity (an/n) = 0 + 0 + ... + 0 = 0.
Therefore, we can conclude that lim n->infinity (an/n) = 0, which means that the sequence (bn) converges to 0.
In conclusion, the theorem states that if the original sequence (an) converges to a, then the sequence (bn), which represents the average of the first n terms of (an), also converges to a.