How to solve $\infty \sum n = 1 \frac{1}{(3 n - 2) (3 n + 1)}$ $sum_(n=1)^oo 1/((3n-2)(3n+1))$ using partial fractions?

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How to solve $\infty \sum n = 1 \frac{1}{(3 n - 2) (3 n + 1)}$ $sum_(n=1)^oo 1/((3n-2)(3n+1))$ using partial fractions?

$\infty \sum n = 1 \frac{1}{(3 n - 2) (3 n + 1)}$ $sum_(n=1)^oo 1/((3n-2)(3n+1))$
There is an infinity sign on the top of the summation sign and n=1 on the bottom.
Please write in details; I don't get how the answer is $\frac{1}{12}$ $1/12$ .

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Answer 1 · 2017-11-25T21:17:20

$\infty \sum n = 1 \frac{1}{(3 n - 2) (3 n + 1)} = \frac{1}{3}$ $sum_(n=1)^oo 1/((3n-2)(3n+1)) = 1/3$

Explanation:

$\frac{1}{(3 n - 2) (3 n + 1)} = \frac{A}{3 n - 2} + \frac{B}{3 n + 1}$ $1/((3n-2)(3n+1)) = A/(3n-2)+B/(3n+1)$

So:

$1 = A (3 n + 1) + B (3 n - 2)$ $1 = A(3n+1)+B(3n-2)$

Putting $n = \frac{2}{3}$ $n=2/3$ this becomes:

$1 = 3 A$ $1 = 3A" "$ so $A = \frac{1}{3}$ $" "A=1/3$

Putting $n = - \frac{1}{3}$ $n=-1/3$ it becomes:

$1 = - 3 B$ $1 = -3B" "$ so $B = - \frac{1}{3}$ $" "B=-1/3$

Then:

$N \sum n = 1 \frac{1}{(3 n - 2) (3 n + 1)} = N \sum n = 1 (\frac{1}{3 (3 n - 2)} - \frac{1}{3 (3 n + 1)})$ $sum_(n=1)^N 1/((3n-2)(3n+1)) = sum_(n=1)^N (1/(3(3n-2))-1/(3(3n+1)))$

$N \sum n = 1 \frac{1}{(3 n - 2) (3 n + 1)} = N \sum n = 1 \frac{1}{3 (3 n - 2)} - N \sum n = 1 \frac{1}{3 (3 n + 1)}$ $color(white)(sum_(n=1)^N 1/((3n-2)(3n+1))) = sum_(n=1)^N 1/(3(3n-2)) - sum_(n=1)^N 1/(3(3n+1))$

$N \sum n = 1 \frac{1}{(3 n - 2) (3 n + 1)} = N \sum n = 1 \frac{1}{3 (3 n - 2)} - N + 1 \sum n = 2 \frac{1}{3 (3 n - 2)}$ $color(white)(sum_(n=1)^N 1/((3n-2)(3n+1))) = sum_(n=1)^N 1/(3(3n-2)) - sum_(n=2)^(N+1) 1/(3(3n-2))$

$color(white)(sum_(n=1)^N 1/((3n-2)(3n+1))) = 1/3+color(red)(cancel(color(black)(sum_(n=2)^N 1/(3(3n-2))))) - color(red)(cancel(color(black)(sum_(n=2)^N 1/(3(3n-2))))) - 1/(3(3N+1)$

$color(white)(sum_(n=1)^N 1/((3n-2)(3n+1))) = 1/3 - 1/(3(3N+1)$

So:

$sum_(n=1)^oo 1/((3n-2)(3n+1)) = lim_(N->oo) sum_(n=1)^N 1/((3n-2)(3n+1))$

$color(white)(sum_(n=1)^oo 1/((3n-2)(3n+1))) = lim_(N->oo) (1/3 - 1/(3(3N+1))) = 1/3$

Note that:

$sum_(n=color(blue)(2))^oo 1/((3n-2)(3n+1)) = 1/(3(3(color(blue)(2))-2)) = 1/12$

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How to solve $\infty \sum n = 1 \frac{1}{(3 n - 2) (3 n + 1)}$ $sum_(n=1)^oo 1/((3n-2)(3n+1))$ using partial fractions?

$\infty \sum n = 1 \frac{1}{(3 n - 2) (3 n + 1)}$ $sum_(n=1)^oo 1/((3n-2)(3n+1))$
There is an infinity sign on the top of the summation sign and n=1 on the bottom.
Please write in details; I don't get how the answer is $\frac{1}{12}$ $1/12$ .

1 Answer

Explanation:

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sum_(n=1)^oo 1/((3n-2)(3n+1))∞∑n=11(3n−2)(3n+1)sum_(n=1)^oo 1/((3n-2)(3n+1)) There is an infinity sign on the top of the summation sign and n=1 on the bottom. Please write in details; I don't get how the answer is 1/121121/12.

1 Answer

Explanation:

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How to solve $\infty \sum n = 1 \frac{1}{(3 n - 2) (3 n + 1)}$ $sum_(n=1)^oo 1/((3n-2)(3n+1))$ using partial fractions?

$\infty \sum n = 1 \frac{1}{(3 n - 2) (3 n + 1)}$ $sum_(n=1)^oo 1/((3n-2)(3n+1))$
There is an infinity sign on the top of the summation sign and n=1 on the bottom.
Please write in details; I don't get how the answer is $\frac{1}{12}$ $1/12$ .