I have to find the sum of the series $$a_n=\frac{1}{\sqrt n}(1+\frac{1}{\sqrt 2}+\cdots+\frac{1}{\sqrt n})$$ where $n$ tends to infinity. This series can be simplified as $$1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$$ where $n$ tends to infinity. And we know that sum of the infinite series $\frac{1}{n}$ is infinity. It means $a_n$ is a divergent series. But using this formula $$\frac{a_1+a_2+\cdots+a_n}{n}=1$$ Sum of $a=1$ which means a is a convergent series. So which one is correct? What will be the sum? $1$ or $\infty$?
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Do you mean $$a_n= \dfrac{1}{\sqrt n(1+\dfrac{1}{\sqrt 2}+\dfrac{1}{\sqrt 3}+.......+\dfrac{1}{\sqrt n})}$$ ? – Khosrotash Feb 19 '17 at 06:57
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No 1/√n is multiplied by the the other part. The question is (1/√n)(1+1/√2+.....+1/√n) Not1/[√n(1+1/√2+.....+1/√n)] – Sona23 Feb 19 '17 at 07:04
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I changed it to latex. Please check if this is correct. – miracle173 Feb 19 '17 at 07:39
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When you say "sum of the series", do you really mean "limit of the sequence"? – TonyK Feb 19 '17 at 11:34
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The first few terms in your simplification should be $\dfrac{1}{\sqrt n}+\dfrac{1}{\sqrt {2n}}+\dfrac{1}{\sqrt {3n}}+\cdots$, not $1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots$ – TonyK Feb 19 '17 at 11:36
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I just mulitplied 1/√n to to the last term 1/√n which became 1/n and then I wrote the series which is 1+1/2+....+1/n – Sona23 Feb 20 '17 at 05:59
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See also: How to find a limit of this sequence: $\lim\limits_{n \to \infty} \sum\limits_{k=1}^n \frac{1}{\sqrt{kn}}$, What is the value of $\lim\limits_{n\to \infty}\left(\sum\limits_{i=1}^n{ \frac{1}{\sqrt{i \cdot n}} } \right)$?, How to evaluate the sum $\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{2n}}+\cdots+\frac{1}{\sqrt{n^2}}$ when $n$ grows? – Martin Sleziak May 10 '20 at 13:11
2 Answers
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You have $n$ terms, and limit for large $n$. Try to convert it to a definite integral.
$$\lim_{n\to\infty}\frac1n\left[f(\alpha_1)+...+f(\alpha_n)\right]$$
Empy2
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I think this question depends on the definition of the sum of a series. In Calculus, we use the definition given by Cauchy, namely the limit of its partial sum. But there are other general definitions like Cesàro sum and Abel sum. You can refer to Wikipedia divergent series
The book entitled Tauberian theory:a century of developments by Jacob Korevarr may be of assistance.
spiritfire
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I suspect the OP means the limit of the sequence, not the sum of the series. – TonyK Feb 19 '17 at 11:33
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@TonyK Limit of a sequence and sum of a series are essentially the same. As you notice, the title is "Sum of the series ...". The problem is how you define this sum. Actually what Sona23 gave were the Cauchy sum and the Cesàro sum. If the Cauchy sum is convergent, then these two coincide by Stolz-Cesàro theorem. In some cases where the Cauchy sum diverges, the Cesàro sum may still exist. – spiritfire Feb 19 '17 at 14:19
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Well, if the OP really means $\lim_{N\rightarrow \infty}\Sigma_{n=1}^Na_n$, then the series is trivially divergent, because $a_n\ge 1$ for all $n$. So I think they must mean $\lim_{n\rightarrow\infty}a_n$, which is not the sum of a series. But in any case their simplification is wrong. I have asked the OP to clarify. – TonyK Feb 19 '17 at 15:23