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The sequence：$$ x_n=1+\frac{1}{\sqrt{2}}+\cdots \cdots +\frac{1}{\sqrt{n}}-2\sqrt{n} $$ My question：$$ x_{n+1}-x_n=\frac{1}{\sqrt{n+1}}-2\sqrt{n+1}+2\sqrt{n} =\frac{2\sqrt{n^2+n}-2n-1}{\sqrt{n+1}}<0 $$ So, the sequence is monotonically decreasing. But, $$ \lim_{n\rightarrow \infty} 1+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{n}}=\lim_{n\rightarrow \infty} \sum_{i=1}^n{\frac{1}{\sqrt{i}}=\lim_{n\rightarrow \infty} \frac{\sqrt{n}}{n}\sum_{i=1}^n{\frac{1}{\sqrt{\frac{i}{n}}}=\lim_{n\rightarrow \infty} \sqrt{n}\int_0^1{x^{-\frac{1}{2}}}dx}}=\underset{n\rightarrow \infty}{\lim}2\sqrt{n} $$ Does that mean that the sum of the previous series is similar to $$ 2\sqrt{n} $$？ So I can think of the limit of this sequence as zero. That's clearly not true given the monotonicity that i had before.

shinomiya
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  • I sort of know that the sum is going to be different than I thought, but I'd like to hear suggestion. – shinomiya Aug 12 '21 at 12:33
  • Not clear to me what the question is. But to check convergence you might first try to get a feel by calculating a couple of terms, at least $x_1$, $x_2$ and $x_3$. – CopyPasteIt Aug 12 '21 at 12:46
  • The right notation to use would be asymptotic ; i.e, $1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{x}}\sim 2\sqrt{x}$ – Mourad Aug 13 '21 at 09:48

2 Answers2

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Note that monotonicity for a sequence $a_n$ doesn't imply that the limit can't be equal to zero (or to any other value).

For example, $a_n = \frac 1 n$ is monotonically decreasing but $a_n \to 0$.

Refer also to the related:

user
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The statement

$$ \lim_{n\rightarrow \infty} 1+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{n}}=\underset{n\rightarrow \infty}{\lim}2\sqrt{n} $$

is useless because is the exact same as saying $\ \infty = \infty,\ $ which tells you nothing. You may as well have written:

$$ \lim_{n\rightarrow \infty} 1+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{n}}=\underset{n\rightarrow \infty}{\lim}2^n\ (= +\infty) $$

Adam Rubinson
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