I have come across a mathematical problem that is to evaluate the expression: $$ lim_{n\rightarrow\infty} \left\{\frac{1}{\sqrt{2n-1^2}}+\frac{1}{\sqrt{4n-2^2}}+\frac{1}{\sqrt{6n-3^2}}+...+\frac{1}{\sqrt{2n^2-n^2}}\right\}\tag1$$ and I have considered the following approach: $$ y=lim_{n\rightarrow\infty} \left\{{\frac{1}{n}}\left[\frac{1}{\sqrt{\frac{2}{n}-\frac{1}{n^2}}}+\frac{1}{\sqrt{\frac{4}{n}-\frac{4}{n^2}}}+\frac{1}{\sqrt{\frac{6}{n}-\frac{9}{n^2}}}+...1\right]\right\} \tag2$$ $$ y=lim_{n\rightarrow\infty}\left\{\Sigma_{i=1}^{n}\frac{1}{\sqrt{\frac{2}{i}-\frac{1}{i^2}}}\frac{1}{i}\right\} \tag3$$ $$ y=\int_0^1f(x)dx \tag4$$ where it is considered that $$f(x)=\frac{1}{\sqrt{2x-x^2}}\tag5$$ and $a=0$ while $b=1$. Integrating thus, we get: $$y=\int_0^1\frac{dx}{\sqrt{2x-x^2}}=[sin^{-1}(x-1)]_0^1=\pi/2\tag6$$ Therefore, the integration says that the sum up to infinite series is $\pi/2$. But I have a query here, which is, if I write the terms of the series, this is what I get here: $$ y=lim_{n\rightarrow\infty}\left\{{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}}+...\frac{1}{n}\right\} \tag7$$ Isn't this a harmonic series? But if so, harmonic series tend to diverge, then how can the infinite sum above be integrated and summed up to a definite value as it has converged? I am being confused.
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$\begingroup$ Please explain how you get $$y=\lim_{n\rightarrow\infty}\left\{{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}}+...\frac{1}{n}\right\}$$ $\endgroup$– GEdgarCommented Jan 13 at 16:54
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$\begingroup$ @GEdgar As I put $n=1, 2, 3$ in the series, where $n$ is the term number, I get that result. For the first term, I considered $n=1$, for the second, it was $n=2$ and so it went on. $\endgroup$– M.RiyanCommented Jan 13 at 17:03
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$\begingroup$ You put $n=1,2,3$ in "the series". Which series? I put equation numbers on your post to help you answer. $\endgroup$– GEdgarCommented Jan 13 at 17:14
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$\begingroup$ @GEdgar At equation $1$. $\endgroup$– M.RiyanCommented Jan 13 at 17:16
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2$\begingroup$ I got it. When $n=1$, the first term in $(1)$ is $1$. When $n=2$, the second term is $1/2$. When $n=3$ the third term is $1/3$. But adding these together is not a meaningful thing to do. $\endgroup$– GEdgarCommented Jan 13 at 17:24
2 Answers
Your summation notation use is not correct which lead to confusion, $$\frac{1}n\cdot n\left(\frac{1}{\sqrt{2n-1^2}}+\frac{1}{\sqrt{4n-2^2}}+\frac{1}{\sqrt{6n-3^2}}+...+\frac{1}{\sqrt{2n^2-n^2}}\right)=\frac{1}n\sum_{i=1}^{n}\frac{1}{\sqrt{\frac{2i}{n}-\frac{i^2}{n^2}}}$$
see $i$ varies but $n$ is constant.
Your mistake,
$$\frac{1}{\sqrt{2n-1^2}}+\frac{1}{\sqrt{4n-2^2}}+\frac{1}{\sqrt{6n-3^2}}+...+\frac{1}{\sqrt{2n^2-n^2}}\ne{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}}+...\frac{1}{n}$$
Because, you cannot put the value of $n$ in $n$th term and up add to get the of the series.
For example, $$\frac{1}{n}(1+2+3+\ldots+n)=\frac{1}n+\frac{2}n+\frac{3}n+\frac{4}n+\ldots+\frac{n}n\ne1+1+1+\ldots n \text{ times}=n$$
As noticed each finite sum has exactly $n$ terms that is:
- $S_1=\frac{1}{\sqrt{2\cdot 1-1^2}}=1$
- $S_2=\frac{1}{\sqrt{2\cdot 2-1^2}}+\frac{1}{\sqrt{4\cdot 2-2^2}}=\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}$
- $S_3=\frac{1}{\sqrt{2\cdot 3-1^2}}+\frac{1}{\sqrt{4\cdot 3-2^2}}+\frac{1}{\sqrt{6\cdot 3-3^2}}=\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{8}}+\frac{1}{\sqrt{9}}$
- $S_4=\frac{1}{\sqrt{2\cdot 4-1^2}}+\frac{1}{\sqrt{4\cdot 4-2^2}}+\frac{1}{\sqrt{6\cdot 4-3^2}}+\frac{1}{\sqrt{8\cdot 4-4^2}}=\frac{1}{\sqrt{7}}+\frac{1}{\sqrt{12}}+\frac{1}{\sqrt{15}}+\frac{1}{\sqrt{16}}$
- $S_5=\frac{1}{\sqrt{2\cdot 5-1^2}}+\frac{1}{\sqrt{4\cdot 5-2^2}}+\frac{1}{\sqrt{6\cdot 5-3^2}}+\frac{1}{\sqrt{8\cdot 5-4^2}}+\frac{1}{\sqrt{10\cdot 5-5^2}}=\frac{1}{\sqrt{9}}+\frac{1}{\sqrt{16}}+\frac{1}{\sqrt{21}}+\frac{1}{\sqrt{24}}+\frac{1}{\sqrt{25}}$
- $\ldots$
which is a completely different pattern with respect to the harmonic series.
