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Let's consider the series $\sum a_n$, with $$a_n = (-1)^{\lfloor \sqrt{n-1}\rfloor} \frac 1 n$$

It looks absolutely like an example for the Leibniz test but here the signs don't interchange one for one. How can we prove the convergence of this series?

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    $\begingroup$ This boils down to assessing the parity of $\lfloor \sqrt{n-1}\rfloor$ . Have you looked in that direction ? $\endgroup$ – Gabriel Romon Feb 18 '16 at 14:51
  • $\begingroup$ little curious to know what is the limit rather than just proving it is convergent. possible? $\endgroup$ – GA316 Feb 18 '16 at 16:11
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You can reduce the question to a question about Liebniz series by inserting parenthesis around the terms that have the same sign. The original series $\sum_{n=1}^{\infty} a_n$ is:

$$ \frac{1}{1} - \frac{1}{2} - \frac{1}{3} - \frac{1}{4} + \frac{1}{5} + \dots + \frac{1}{9} - \frac{1}{10} - \dots. $$

Denote the partial sums of the series $\sum_{n=1}^{\infty} a_n$ by $S_n = \sum_{k=1}^n a_k$. Inserting parenthesis around the terms with the same sign, we obtain a new series $\sum_{n=1}^{\infty} b_n$:

$$ \frac{1}{1} - \left( \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \right) + \left( \frac{1}{5} + \dots + \frac{1}{9} \right) - \dots $$

Formally, what we do is we define $b_n$ by letting

$$ c_n = \sum_{k=(n-1)^2 + 1}^{n^2} \frac{1}{k}, \\ b_n = (-1)^{n+1} c_n = (-1)^{n+1} \sum_{k=(n-1)^2 + 1}^{n^2} \frac{1}{k}. $$

Denote by $T_n$ the partial sums of the series $\sum_{n=1}^{\infty} b_n$. The fact that the new series $\sum_{n=1}^{\infty} b_n$ is obtained by inserting parenthesis around the terms corresponds to the fact that the sequence of partial sums of $\sum_{n=1}^{\infty} b_n$ is a subsequence of the sequence of partial sums of $\sum_{n=1}^{\infty} a_n$. Namely, $T_n = S_{n^2}$. If we can show that $c_n$ are monotonically decreasing to zero, we will obtain that $\sum_{n=1}^{\infty} b_n$ is a Liebniz series and hence converges. This shows that $S_{n^2}$ has a limit. Since the series $\sum_{n=1}^{\infty} b_n$ was obtained from the series $\sum_{n=1}^{\infty} a_n$ by inserting parenthesis around terms that have the same sign, this implies that $S_n$ also has a limit (the same limit).

Estimating $c_n$, we have

$$ c_n = \sum_{k=(n-1)^2 + 1}^{n^2} \frac{1}{k} \leq \sum_{k=(n-1)^2 + 1}^{n^2} \frac{1}{(n-1)^2 + 1} = \frac{n^2 - (n-1)^2}{(n-1)^2 + 1} = \frac{2n - 1}{(n-1)^2 + 1} \xrightarrow[n \to \infty]{} 0 $$

so $c_n$ tends to zero.

Finally,

$$ c_n - c_{n+1} = \sum_{k=(n-1)^2 + 1}^{n^2} \frac{1}{k} - \sum_{k=n^2 + 1}^{(n+1)^2} \frac{1}{k} \\ = \sum_{i = 0}^{2n - 2} \left( \frac{1}{(n-1)^2 + 1 + i} - \frac{1}{n^2 + 1 + i} \right) - \frac{1}{(n+1)^2 - 1} - \frac{1}{(n+1)^2} \\ = \sum_{i = 0}^{2n - 2} \frac{2n - 1}{\left( (n-1)^2 + 1 + i \right) \left( n^2 + 1 + i \right)} - \frac{1}{(n+1)^2 - 1} - \frac{1}{(n+1)^2} \\ \geq \frac{(2n-1)^2}{\left( (n-1)^2 + 1 + (2n-2) \right) \left( n^2 + 1 + (2n-2) \right)} - \frac{1}{(n+1)^2 - 1} - \frac{1}{(n+1)^2} \\ \geq \frac{(2n-1)^2}{n^2(n^2 + 2n - 1)} - \frac{2}{(n^2 + 2n - 1)^2} \\ = \frac{2(n-1)^2 - 1}{(n^2 + 2n + 1)^2} > 0 $$

for $n > 1$ which shows that $c_n$ are monotonically decreasing.

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  • $\begingroup$ do we know the limit also? simply curious to know. $\endgroup$ – GA316 Feb 18 '16 at 16:11
  • $\begingroup$ @GA316 That's a good question - I don't know how to compute the limit. $\endgroup$ – levap Feb 18 '16 at 16:17
  • $\begingroup$ @levap can you point me to a proof that since we gathered terms around the same sign, it implies that $S_n$ has the same limit? $\endgroup$ – marmistrz Feb 22 '16 at 10:29
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    $\begingroup$ We teach this in our calculus course but I don't know a written reference. In your case, you have $S_{g(n)^2} \leq S_n \leq S_{f(n)^2}$ where $f,g$ are not monotonic but tend to infinity as $n$ goes to infinity. For example, $S_4 \leq S_2 \leq S_1$, $S_4 \leq S_6 \leq S_9$, etc. This also happens in the general case. $\endgroup$ – levap Feb 22 '16 at 13:38
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    $\begingroup$ By not monotonic I mean not strictly increasing but only weakly monotonic. $\endgroup$ – levap Feb 22 '16 at 14:03

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