Question

Find the sum of the series sigma n-infinity (4/(4n-3) (4n+1))

Answer 1

SOLUTION

TO DETERMINE

The sum of the series

$\displaystyle \sf{\sum\limits_{n=1}^{ \infty } \: \frac{4}{(4n - 3)(4n + 1)} }$

CONCEPT TO BE IMPLEMENTED

The infinite series

$\displaystyle \sf{\sum\limits_{n=1}^{ \infty } \:u_n }$

is said to be convergent or divergent according as the sequence $\displaystyle \sf\{ \: s_n \}$ is convergent or divergent

Where $\displaystyle \sf{\: s_n = u_1 + u_2 + .... +u_n }$

$\displaystyle \sf\{ \: s_n \}$ is called partial sums of the series $\displaystyle \sf{ \sum \: u_n }$

If $\displaystyle \sf{ lim \: s_n = s}$

Then s is called sum of the series $\displaystyle \sf{ \sum \: u_n }$

EVALUATION

Here the given series is

$\displaystyle \sf{\sum\limits_{n=1}^{ \infty } \: \frac{4}{(4n - 3)(4n + 1)} }$

Let

$\displaystyle \sf{\sum\limits_{n=1}^{ \infty } \:u_n }$

be the given series

Then

$\displaystyle \sf{ \:u_n = \frac{4}{(4n - 3)(4n + 1)} }$

$\therefore \: \displaystyle \sf{ \:u_n = \frac{(4n + 1) - (4n - 3)}{(4n - 3)(4n + 1)} }$

$\therefore \: \displaystyle \sf{ \:u_n = \frac{1}{4n - 3} - \frac{1}{4n + 1} }$

Now $\displaystyle \sf{\: s_n = u_1 + u_2 + .... +u_n }$

$\displaystyle \sf{\implies \: s_n = 1 - \frac{1}{5} + \frac{1}{5} - \frac{1}{13} + \frac{1}{13} - ..... - \frac{1}{4n + 1} }$

$\displaystyle \sf{\implies \: s_n = 1 - \frac{1}{4n + 1} }$

$\therefore \: \: \displaystyle \sf{ lim \: s_n = 1}$

Hence the given series is convergent and the sum of the series is 1

FINAL ANSWER

The sum of the given series = 1

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