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If I can show that $\displaystyle \sum_{k\geq1}a_k^2<\infty$, then is this an acceptable proof that $\displaystyle \sum_{k\geq1}a_k^2$ is convergent?

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That is not enough. Use the definition of convergence. –  Quadrescence Mar 20 '11 at 0:31

2 Answers 2

up vote 4 down vote accepted

By defintion, $\sum\limits_{k\geq 1}b_k$ converges if and only if the sequence of partial sums converges, $\lim\limits_{n\to\infty} s_n = L$, where $s_n=\sum\limits_{k=1}^n b_k$.

If $b_k\geq 0$ for all $k$, then the sequence of partial sums is monotone, $s_1\leq s_2\leq s_3\leq\cdots$.

A monotone sequence converges if and only if it is bounded. Therefore, for a series of nonnegative terms $b_k\geq 0$, we have that $\sum\limits_{k\geq 1}b_k$ converges if and only if the sequence $s_n = \sum\limits_{k=1}^nb_k$ is bounded, if and only if $\lim\limits_{n\to\infty}s_n\lt\infty$, if and only if $\sum\limits_{k\geq 1}b_k\lt \infty$.

That is: for a series with nonnegative terms, the only two possibilities are that the series is bounded and converges, or that the series diverges to $\infty$.

If $b_k = a_k^2$ and all $a_k$ are real, then $b_k\geq 0$, so the observation above applies: if you can prove that the partial sums are bounded (which is the only reasonable meaning of $\sum\limits_{k\geq 1}a_k^2\lt \infty$), then the series necessarily converges.

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It is true if $a_k$ is eventually nonnegative (there exists $K$ such that $a_k \geq 0$ for $k > K$.

The series $\sum_{k \geq 1} (-1)$ satisfy your inequality and diverges.

If it is the case where $a_k \geq 0$ then the sequence $S_n := \sum_{1 \leq k \leq n} a_k$ is eventually increasing and bounded. Hence it converges.


For the edited question, yes, it is true. Your inequality is now equivalent to the convergence of the series. The reason is as above. The sequence of partial sums of your series is monotone increasing and bounded, hence converges.

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I think you mean $a_k > 0$ for some non-finite subsequence of $a$. –  Quadrescence Mar 20 '11 at 0:35
    
Edited the question. –  Vafa Khalighi Mar 20 '11 at 0:39
    
Edited the answer. It is quite similar as to the previous question. –  xen Mar 20 '11 at 0:49

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