If $\sum_{i=0}^\infty a_n^2$ is convergent, does that imply $\sum_{i=0}^\infty a_n$ is convergent?
How can I prove that?
Thanks for the help.
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8No. $a_n=1/n$... – Mark Viola Oct 03 '15 at 04:42
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$1+1/2+1/3+1/4+1/5+1/6+17+1/8+...... \ge 1+1/2+(1/4+1/4)+(1/8+1/8+1/8+1/8)+... =1+1/2+1/2+1/2+...=\infty$, while for$ n \ge 2$, we have $\sum_{j=2}^{j=n}< \sum_{j=2}^{j=n} (j(j-1))^{-1}=\sum_{j=1}^{j=n} (1/(j-1)-1/j=1-1/n$ – DanielWainfleet Oct 03 '15 at 04:56
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|Are you familiar with the Cauchy condensation test? If $A(n) /ge A(n+1) \ge 0$ then $\sum_n A(n)$ converges iff $\sum_n 2^n A(2^n)$ converges – DanielWainfleet Oct 03 '15 at 05:04
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1The converse would be a slightly more interesting question. – abligh Oct 03 '15 at 08:13
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We have the counterexample $a_n=\frac1n$. The series $\sum_{n=1}^\infty \frac1{n^2}=\frac{\pi^2}{6}$ converges. But the harmonic series $\sum_{n=1}^\infty \frac1{n}$ is divergent.
Mark Viola
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As mentioned in other answers, $a_k=\frac1n$ is a counterexample, since $\sum_{k=1}^\infty \frac1{n}$ diverges and $\sum_{k=1}^\infty \frac1{n^2}$ converges. If you want to see proofs of these facts, you can look here:
- Why does the series $\sum_{n=1}^\infty\frac1n$ not converge? and other questions linked there
- Need to prove the sequence $a_n=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}$ converges and other questions linked there
- If you are also interested in the value of the sum $\sum_{k=1}^\infty \frac1{n^2}$: Different methods to compute $\sum\limits_{k=1}^\infty \frac{1}{k^2}$ and other questions linked there
The implication in the opposite direction is true for $a_k>0$, i.e., for series with positive terms. See Prove that if $\sum{a_n}$ converges absolutely, then $\sum{a_n^2}$ converges absolutely and other questions linked there.
Martin Sleziak
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take $a_n=1/n$
then $\sum \frac{1}{n^2}$ is convergent while $\sum \frac{1}{n}$ is divergent
user300
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