Is it true that if $\alpha_k\to0$ and $\sum\limits_{k=0}^\infty\alpha_k=\infty$,

we have $\sum\limits_{k=0}^\infty(\alpha_k)^2= c$ (some constant) or at least it does not converge to $\infty$?

  • $\begingroup$ @BigM I think $k$ is just the index of the sequence here $\endgroup$ – fonini Feb 10 '15 at 15:30
  • $\begingroup$ Oh alright then. $\endgroup$ – BigM Feb 10 '15 at 15:31
  • $\begingroup$ @lolibility I think you mean $\sum (a^k)^2<\infty$ ? There's no reason for the sum to be zero. $\endgroup$ – fonini Feb 10 '15 at 15:31
  • $\begingroup$ (I'm not sure if the sum of the squares converges indeed or not, and I actually think it doesn't) $\endgroup$ – fonini Feb 10 '15 at 15:33
  • $\begingroup$ Your notation is flawed. You shouldn't write $\sum t_k \to c$, but rather $\sum t_k = c$. That's because the upper limit of the sum is $\infty$, which means the expression is already a limit. If you had written partial sums with a finite upper index on the sum, then the "$\to$" would be appropriate. Also, your upper index $k$ on the term is confusing -- it looks like an exponent. $\endgroup$ – MPW Feb 10 '15 at 15:37

Consider the sequence $α_k = \frac{1}{\sqrt{k}}$ this sequence converges against $0$, the sum $\sum_{k=0}^∞ α_k$ diverges and the sum of squares $\sum_{k=0}^∞ α_k^2$ diverges as well.

If that is unclear, have a look at the harmonic series.


No. a handy typical example: let $\alpha_k=\frac{1}{k}$, then $\alpha_k\to 0$ and $\sum\limits_{k=0}^\infty\alpha_k=\infty$. But $\sum\limits_{k=0}^\infty\alpha_k^2=\frac{\pi^2}{6}$ Side note :unless all $\alpha_k$ are zero, the series of squares ,if converges, it does not converge to zero.

  • $\begingroup$ This is an example of the case the OP thinks is universal although it is not. $\endgroup$ – Did Feb 10 '15 at 15:46

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