Weird sum of series

Calculus Level 5

$\large{\frac{1}{1\times 7}+\frac{1}{2\times 9}+\frac{1}{3\times 11}+\frac{1}{4\times 13}+\ldots }$

The value of above series is ${\frac{A}{B}-\frac{C}{D}\ln(2)}$ where $A,B,C,D,$ are positive integers.

Find minimum value of $A+B+C+D+2$ .

Try my set .

The answer is 130.

2 solutions

Sudeep Salgia
Jul 25, 2015

$\displaystyle S = {\frac{1}{1\times 7}+\frac{1}{2\times 9}+\frac{1}{3\times 11}+\frac{1}{4\times 13}+........} = \sum_{r=1}^{\infty } \frac{1}{r(2r+5)}$

$\displaystyle \frac{5S}{2} = \sum_{r=1}^{\infty } \frac{5}{2r(2r+5)} = \sum_{r=1}^{\infty } \frac{1}{2r} - \frac{1}{(2r+5)}$

$\displaystyle \begin{array}{c}\\ \frac{5S}{2} && = \sum_{r=1}^{\infty } \left( \frac{1}{2r} - \frac{1}{(2r-1)} \right) - \left( 1 + \frac{1}{3} + \frac{1}{5} \right) \\ && = 1 + \frac{1}{3} + \frac{1}{5} - \left( 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} \dots \right) \\ && = \frac{23}{15} - \ln (2) \\ \end{array}$

$\displaystyle \therefore S = \frac{46}{75} - \frac{2}{5} \ln (2)$ .

Compare the values and add them to obtain the answer as $\displaystyle \boxed{130}$ .

Moderator note:

With infinite series, you have to be very careful with rearranging of terms especially when we only have a conditionally convergent sequence. For example, this statement is not true:

$\frac{5S}{2} = \left( \frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \dots \right) - \left( \frac{1}{7} + \frac{1}{9} + \frac{1}{11} + \dots \right)$

@Calvin Lin I know it is technically incorrect to write it that way. But I have expressed it that way to make the manipulations look clearer and less vague.

Sudeep Salgia - 5 years, 10 months ago

I disagree. It does not make it clearer.

In fact, someone else could say "The terms in the parenthesis now evaluate to $\infty - \infty$ , which is indeterminate". But that is not the answer.

Calvin Lin Staff - 5 years, 10 months ago

I get that. I apologize for the mistake. I have edited it now. Does it look better?

Sudeep Salgia - 5 years, 10 months ago

@Sudeep Salgia – Yes, that's much better.

I haven't found good notation to indicate "We haven't moved the terms by much", which is needed for this conditional convergence.

Calvin Lin Staff - 5 years, 10 months ago

Tanishq Varshney
Jul 25, 2015

We know

$\large{\ln(1-x)=-(x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+...)}$

$\large{x^4 \ln(1-x^2)=-(x^6+\frac{x^8}{2}+\frac{x^{10}}{3}+....)}$

integrating

$\large{\displaystyle \int^{1}_{0} x^4 \ln(1-x^2)dx=-(\frac{x^7}{1\times7}+\frac{x^9}{2\times 9}+\frac{x^{11}}{3\times 11}+...)_{0}^{1}}$

apply integration by parts and solve the integral

$\large{\displaystyle \int^{1}_{0} x^4 \ln(1-x^2)dx=-\frac{46}{75}+\frac{2}{5} \ln(2)}$

You can also use the result that $H_n\sim \ln(n)+\gamma$ when $n\to\infty$ where $H_n$ is the $n^{\textrm{th}}$ harmonic number and $\gamma$ is Euler-Mascheroni constant. With a bit of manipulation, we get,

$S=\sum_{k=1}^\infty\frac{1}{k(2k+5)}=\frac 15\lim_{n\to\infty}\left(H_n-2\left(H_{5+2n}-H_6-\frac 12(H_{n+2}-H_3)\right)\right)$

Applying the stated result and calculating the finite terms and dividing the numerator and denominator of the fraction inside the logarithm by $n^2$ , you easily get,

$S=\frac 15\left(\frac{46}{15}+\lim_{n\to\infty}\ln\left(\frac{1+2/n}{4+20/n+25/n^2}\right)\right)=\frac{46}{75}-\frac 25\ln(2)$

Prasun Biswas - 5 years, 10 months ago

Super! Didn't think of this method, I used partial fraction and regrouping to reach the answer...

Kishore S. Shenoy - 5 years, 10 months ago

Weird sum of series

Try my set .

The answer is 130.

2 solutions

Moderator note:

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