If the series $ \ \sum_{n=1}^{\infty} a_n^2 \ $ converges , then the $ \ n^{th } \ $ term $ \ a_n^2 \ \to 0 \ \ as \ \ n \to \infty $

Question

(i) If the series $ \ \sum_{n=1}^{\infty} a_n^2 \ $ converges , then the $ \ n^{th } \ $ term $ \ a_n^2 \ \to 0 \ \ as \ \ n \to \infty $.

then show that $ \ a_n \to 0 \ \ as \ \ n \to \infty \ $

(ii) If the series $ \ \sum_{n=1}^{\infty} a_n^3 \ $ converges , then the $ \ n^{th } \ $ term $ \ a_n^3 \ \to 0 \ \ as \ \ n \to \infty $.

then show that $ \ a_n \to 0 \ \ as \ \ n \to \infty \ $

Answer:

(i)

I am thinking this holds trivially. Because $ \ a_n^2 \to 0 \ \Rightarrow a_n \to 0 \ $

But I can not prove it mathematically.

I need clear explanation .

help me

Answer 1

Let $\epsilon>0$ be given. Then $\epsilon^k>0$. Hence $a_n^k\to 0$ tells us that there exists $N$ such that $|a_n^k|<\epsilon^k$ for all $n>N$. Then for such $n$, we also have $|a_n|<\epsilon$ (because $x\ge y>0$ implies $x^k\ge y^k$, and of course $|x^k|=|x|^k$).

We conclude that $a_n\to 0$.

Answer 2

$$ 0 = \lim a_n^2 = (\lim a_n ) (\lim a_n ) \implies \lim a_n = 0 $$

Answer 3

Apply the Cauchy criterion of convergence to the partial sums $s_N = \sum_{n=0}^Na_n^2$ to see that $a_n^2 \to 0$: $$ a_{N+1}^2 = \sum_{n=N}^{N+1}a_n^2 < \epsilon \ \text{for all $N$ sufficiently large.} $$ Use continuity of $\sqrt{(\cdot)}\colon\Bbb R_{\ge0}\to\Bbb R_{\ge0}$ to conclude that $\lim_{n\to\infty}|a_n| =\lim_{n\to\infty}\sqrt{a_n^2}= \sqrt{\lim_{n\to\infty}a_n^2} = \sqrt{0} = 0$. Since $|a_n|\to 0$, use the definition of convergence to conclude that $a_n\to 0$.

For cubes, use the exact same strategy except take cube roots $\sqrt[3]{(\cdot)}$.

Answer 4

$a_n^2 \rightarrow 0$ means for every $\epsilon>0$, there is an $N_{\epsilon} \in \Bbb{N}$ so that $$\vert a_n^2-0 \vert <\epsilon$$ for all $n>N_{\epsilon}$.

So that $$\vert a_n \vert^2 <\epsilon$$ for all $n>N_{\epsilon}$ and so $$\vert a_n \vert <\sqrt{\epsilon}$$ for all $n>N_{\epsilon}$

The latter inequality shows $a_n \rightarrow 0$ as well!