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The sum of series

$$\frac{1}{1\cdot 3}+\frac{1}{4\cdot 5}+\frac{1}{7\cdot 7}+\frac{1}{10\cdot 9}+\cdots $$

My attempt $$\displaystyle \sum^{n}_{k=1}\frac{1}{(3k-2)(2k+1)}=\frac{1}{7}\sum^{n}_{k=1}\bigg[\frac{3}{3k-2}-\frac{2}{2k+1}\bigg]$$

$$=\frac{1}{7}\sum^{n}_{k=1}\int^{1}_{0}\bigg(3x^{3k-3}-2x^{2k}\bigg)dx$$

$$=\frac{1}{7}\int^{1}_{0}\bigg(\sum^{n}_{k=1}3x^{3k-3}-2x^{2k}\bigg)dx$$

$$=\frac{1}{7}\int^{1}_{0}\bigg[\frac{3(1-x^{3n})}{1-x^3}-\frac{2(1-x^{2n+2})}{1-x^2}\bigg]dx$$

How do i solve it Help me please

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    $\begingroup$ Why do you sum only till finite n. Your title suggests an infinite sum. $\endgroup$ Jun 21, 2019 at 8:33
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    $\begingroup$ $$ \sum_{k=1}^n x^{2k} = \frac{x^2(1-x^{2n})}{1-x^2}$$ $\endgroup$ Jun 21, 2019 at 8:41
  • $\begingroup$ I've noticed something strange here, when trying to use Mathematica to check my results. I get a different result for $$ \sum_{k=1}^\infty \int_0^1 (3 x^{3 k - 3} - 2 x^{2 k}) dx$$ than for $$ \int_0^1 \big(\sum_{k=1}^\infty (3 x^{3 k - 3} - 2 x^{2 k}) \big)dx$$ anyone can explain why? The integral and the series seem convergent enough. Or am I wrong? $\endgroup$ Jun 21, 2019 at 10:53

3 Answers 3

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Such integrals can be expressed by the digamma function. One of the formulae for the digamma function is

$$ \psi(z+1) = -\gamma + \int_0^1 \frac{1-t^z}{1-t}dt $$

You have then $$ \int_0^1 \frac{1-x^{3n}}{1-x^3} dx = \int_0^1 \frac{1-t^{n}}{1-t} \frac13t^{-\frac23}dt = \int_0^1 \frac13 \Big(\frac{1-t^{n-\frac23}}{1-t}-\frac{1-t^{-\frac23}}{1-t}\Big)dt = \frac13\big(\psi(n+\frac13)-\psi(\frac13)\big)$$ $$ \int_0^1 \frac{x^2(1-x^{2n)}}{1-x^2} dx = \int_0^1 \frac{t(1-t^{n})}{1-t} \frac12t^{-\frac12}dt = \int_0^1 \frac12 \Big(2\frac{1-t^{n+\frac12}}{1-t}-2\frac{1-t^{\frac12}}{1-t}\Big)dt = \frac12\big(\psi(n+\frac32)-\psi(\frac32)\big)$$ so $$ \sum_{k=1}^n \frac{1}{(3k-2)(2k+1)} = \frac17\big(\psi(n+\frac13)-\psi(n+\frac32)-\psi(\frac13)+\psi(\frac32)\big)$$ The same formula can be derived faster, using the fact that $$ \psi(n+z)-\psi(z) = \sum_{k=0}^{n-1}\frac{1}{z+k}$$ so $$ \sum_{k=1}^{n}\frac{1}{k-\frac23}=\sum_{k=0}^{n-1}\frac{1}{k+\frac13} = \psi(n+\frac13)-\psi(\frac13) $$ $$ \sum_{k=1}^{n}\frac{1}{k+\frac12}=\sum_{k=0}^{n-1}\frac{1}{k+\frac32} = \psi(n+\frac32)-\psi(\frac32) $$

If you want to calculate the infinite sum, you have \begin{align} \sum_{k=1}^\infty \frac{1}{(3k-2)(2k+1)} &= \frac17 \int_0^1 \Big(\frac{3}{1-x^3} - \frac{2x^2}{1-x^2}\Big) dx =\\ &= \frac17\int_0^1 \Big(\big(\frac{1}{1-x}+\frac{x+2}{x^2+x+1}\big) - \big(-2 + \frac{1}{1-x}+\frac{1}{x+1}\big)\Big) dx =\\ &= \frac17 \int_0^1 \Big(\frac{x+2}{x^2+x+1} +2 - \frac{1}{x+1}\Big) dx \end{align} and the last integral can be computed with standard methods for the integrals of rational functions:

\begin{align} \int_0^1 \frac{x+2}{x^2+x+1} dx &= \int_0^1 \frac{(x+\frac12)+\frac32}{(x+\frac12)^2+\frac34} dx =^{t=\frac{2}{\sqrt{3}}(x+\frac12)} \\ &= \int_{1/\sqrt{3}}^{\sqrt{3}} \frac{t+\sqrt{3}}{t^2+1} dt = \\ &= \Big(\frac{1}{2}\log(t^2+1) + \sqrt{3}\arctan t\Big)\Big|_{t=1/\sqrt{3}}^{t=\sqrt{3}} = \\ &= \frac{1}{2}\log 3 + \frac{\pi\sqrt{3}}{6}\end{align} $$ \int_0^1\frac{1}{x+1} dx = \log(x+1)\big|_{x=0}^{x=1}= \log 2 $$ so $$ \sum_{k=1}^\infty \frac{1}{(3k-2)(2k+1)} =\frac17 (2+ \frac{\pi\sqrt{3}}{6}+\frac{1}{2}\log 3 -\log 2) $$

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(CONSIDERING INFINITE SUM)

We have $$I_n=\frac 37\int_0^1 \frac {1-x^{3n}}{1-x^3} dx-\frac 27\int_0^1 \left(\frac {1-x^{2n+2}}{1-x^2} +\frac {x^2-1}{1-x^2}\right) dx$$

Let the first integral be denoted by $I_{1,n}$ and the second be denoted by $I_{2,n}$

Hence for the sum we need $$\lim_{n\to\infty} \frac 37 I_{1,n}-\frac 27 I_{2,n}$$

Now note that using the substitution $x^3=t$ in $I_{1,n}$ we get $$I_{1,n}=\frac 13 \int_0^1 \left(\frac {1-t^{n-\frac 23}}{1-t} -\frac {1-t^{-\frac 23}}{1-t}\right)dt$$

Now using the Integral relation if Digamma function I.e. $$\psi(z+1)+\gamma=\int_0^1 \frac {1-x^z}{1-x}dx$$

We get $$I_{1,n}=\frac 13\left(\psi\left(n+\frac 13\right)-\psi\left(\frac 13\right)\right)$$

Similarly using the substitution $x^2=u$ and the same integral relation in $I_{2,n}$ we get $$I_{2,n}= \frac 12\left(\psi\left(n+\frac 32\right)-\left[\psi\left(\frac 12\right)+\frac {1}{1/2}\right]\right)$$

But using that $$\psi(z+1)=\psi(z)+\frac 1z$$ we have $$I_{2,n}= \frac 12\left(\psi\left(n+\frac 32\right)-\psi\left(\frac 32\right)\right)$$

Now taking the limit we see that $$\lim_{n\to\infty} I_n=\frac 17\left[\lim_{n\to\infty} \left(\psi\left(n-\frac 23\right)-\psi\left(n+\frac 32\right)\right)\right]+\frac {\psi\left(\frac 32\right)-\psi\left(\frac 13\right)}{7}$$

Now for large $n$; $\psi(n)\sim \ln n$ . Using this the limit inside square brackets turns $0$

Thus the sum is equal to $$\frac {\psi\left(\frac 32\right)-\psi\left(\frac 13\right)}{7}$$

Which can be simplified using known values of Digamma function.

Edit: You can use from Wikipedia that $$\psi\left(\frac 13\right)=-\frac {\pi}{2\sqrt 3}-\frac {3\ln 3}{2}-\gamma$$ And $$\psi\left(\frac 32\right)=2-2\ln 2-\gamma$$

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The approximate value of sum of terms in $HP$ is $S_n\approx \frac{1}{d}\ln(\frac{2a+2(n-1)d}{2a-d})$ here $a$=reciprocal of first term $(a_1=4,a_2=3)$ respectively. $d$=difference between the reciprocal of two terms $(d_1=3,d_2=2)$ respectively Note that the two HPs are $3+3(\frac{1}{4}+\frac{1}{7}+....)-2(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...)$

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