If $A_i$ is a sequence such that: $ \displaystyle \lim_{n\to \infty} A_i =A $. How do you prove: $\displaystyle \lim_{n\to \infty}\sum_{i=1}^{n}A_i=\lim_{n\to \infty} nA $
Infinite sum of a convergent sequence, which does not tend to zero
0
$\begingroup$
real-analysis
sequences-and-series
limits
1 Answers
0
If $A > 0$, $A_n > A/2$ for all sufficiently large $n$, so $\sum_{i=1}^n A_i > \text{constant} + nA/2 \to + \infty$. Similarly if $A < 0$.
-
0If you are allowed to make the "sufficiently large $n$" argument, why not just say $A_i=A$ for $\forall i$. So $\sum A_i=nA$. Coming from a physics background it really confuses me what you mathematicians consider necessary to prove. – 2017-02-21
-
0@RuskoRuskov $A_i = A$ for all $i$ is an **example**. One example, or even infinitely many, does not prove a general statement. – 2017-02-21