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As we all know, this series $$\sum_{n=1}^{\infty}\dfrac{1}{n^{\frac{1}{2}}}$$ is divergent. My question is that if there exists an upper bound of this series.

luvtsubasa
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    As all summands are positive, so the sum will become increasingly large when you sum up to some fixed number $N$ and then make $N$ larger. So no, by definition of divergence and positivity of all term no such upper bound exists, unless I misunderstood. – Jonas Lenz Dec 19 '23 at 17:29
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    You mean a nice sequence $(C_n)$ such that $$ \sum_{n=1}^N\frac{1}{\sqrt{n}}\leqslant C_N ? $$ – Tuvasbien Dec 19 '23 at 17:31
  • In case you want the upper-bound sequence as @Tuvasbien pointed out I would suggest observe that the partial sums can be bounded by: one plus the definite integral of $x^{-1/2}$ on the interval $[1,n-1]$. – Leandro Dec 19 '23 at 17:38
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    See https://math.stackexchange.com/q/1211973/42969 for lower and upper bounds of the partial sums. – Martin R Dec 19 '23 at 17:38
  • Denoting $a_N=\sum_{n=1}^N\frac1{\sqrt n}$ and using Stolz–Cesàro theorem (https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem) $$\lim_{N\to\infty}\frac{a_N}{\sqrt N}=\lim_{N\to\infty}\frac{a_{N}-a_{N-1}}{\sqrt{N}-\sqrt {N-1}}=\lim_{N\to\infty}\frac{\frac1{\sqrt N}}{\frac1{\sqrt N+\sqrt{N-1}}}=2$$ – Svyatoslav Dec 19 '23 at 20:27

2 Answers2

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If we call $s_n$ the sequence of partial sums of this series, then $\sum_{n=1}^{\infty} \frac{1}{n^\frac{1}{2}}$ = $\lim_{n \to \infty} s_n$. Since for all $n \in \mathbb{N}$, we have $\frac{1}{n^\frac{1}{2}} > 0$, we have that $s_n$ is a strictly increasing sequence. Now, if $s_n$ was bounded above, then by the principle of bounded monotone convergence, we could conclude that it converges. Clearly, this is a contradiction, since $\lim_{s \to \infty} s_n$ $\to \infty$.

GTBailey
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This is a series of positive terms. If it has an upper bound, then its corresponding sequence of partial sums will be strictly increasing, bounded above and so convergent. This implies series is convergent. So, the series can not have an upper bound.

Math
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