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Let $(a_n)$ be a sequence of positive terms and suppose that $\sum_\limits{n=0}^{\infty} a_n$ converges. Show that $\sum_\limits{n=0}^{\infty} a_n^2$ converges.

This is in the section on the Comparison Test so that must be what I'm supposed to use. But I don't see how. $(a_n)^2$ might be smaller or larger than $a_n$ depending on $a_n$. And I can't use the Comparison Test with some other series because there's no info here about how fast $\sum a_n$ converges. Hmm. Any hints?

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marked as duplicate by Watson, Daniel W. Farlow, Simply Beautiful Art, Namaste calculus Nov 11 '16 at 0:34

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  • $\begingroup$ Note that $\frac{a_n^2}{a_n}\to 0$. $\endgroup$ – A.Γ. Nov 10 '16 at 19:43
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    $\begingroup$ HInt: if $\sum_{n=0}^{\infty} a_n$ converges then $a_n \to 0$ Hence $a_n < 1 $ for $n>M$ for some $M$. $\endgroup$ – leonbloy Nov 10 '16 at 19:44
  • $\begingroup$ You can use the comparison after a certain point. $\endgroup$ – Arthur Nov 10 '16 at 20:15
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If we know that $a_n\to 0$ then it is going to satisfy $0\le a_n<1$ after a while, that is $a_n^2\le a_n$ for $n\ge N$ with some $N>0$. It makes the comparison for the tails valid $$ \sum_{n=N}^\infty a_n^2\le\sum_{n=N}^\infty a_n. $$ Convergence of the tails is all we need for convergence.

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For any $\epsilon>0$, there must be an $N$ such that $a_n<\epsilon$ for all $n>N$. If this were not true, you'd be adding an infinite amount of terms which didn't approach zero, meaning the sum would diverge, which, according to the statement, is false.

Let $\epsilon=1$ so that we have some $N$ such that $a_n<1$ for all $n>N$. It should then be clear that for any $0<a_n<1$, we have $0<(a_n)^2<a_n<1$.

So, we have

$$\sum_{n>N}(a_n)^2<\sum_{n>N}a_n$$

And since $\sum_{n=1}^N(a_n)^2$ is finite, we show by comparison test that

$$\sum_{n\ge0}(a_n)^2<\sum_{n=1}^N(a_n)^2+\sum_{n>N}a_n$$

which converges.

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If $a_n\ge0$ for all $n$ and $\sum_{n=1}^\infty a_n=A$, then $A\ge a_n$ for all $n$, which implies $a_n^2\le Aa_n$. So you can compare $\sum a_n^2$ to $\sum Aa_n=A\sum a_n$.

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Hint: as an alternative way, it's probably worth mentioning that for positive terms: $$\left(\sum_{n=0}^{k} a_n\right)^2 \geq \sum_{n=0}^{k} a_n^2$$ Now, taking the limit $$\left(\sum_{n=0}^{\infty} a_n\right)^2 \geq \sum_{n=0}^{\infty} a_n^2$$

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As the section is about the Comparison Test then the most appropriate is to use the one-sided limit comparison test: $$ \sum_n a_n\ \text{converges and }\frac{a_n^2}{a_n}=a_n\to 0<+\infty\quad\Rightarrow\quad \sum_n a_n^2\ \text{converges.} $$

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$(a_0\,+a_1\,+ \,a_2+.........) (a_0\,+a_1\,+ \,a_2+.........)\,=\,L$, where $L$ is a finite number, which equals $(a_0^2\, +\, a_1^2+\,a _2^2+......)\, +\, a$, where $a$ is a number less than $L$, and the terms are all positive, hence $\sum a_n^2$, converges to a finite number

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take two sequences $x_n$=$\sum_{k=0}^{n}a_k$ and $t_n$=$\sum_{k=0}^{n}(a_k)^2$

now $a_n$> 0 for all n$\in$N so we can write $t_n$ < $(x_n)^2$ for all n. now as $\sum a_n$ is convergent as lim$x_n$ exists.. then the sequence $x_n$ is convergent and so it is bounded ..then there exists M>0 s.t $x_n$< M for all n$\in$N . so $t_n$< $M^2$ for all n$\in$N . now $t_n$ is increasing strickty and bounded above by $M^2$

so $t_n$ is convergent..means $\sum (a_n)^2$ is convergent

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