Question

How do I find the value of this series?

Answer 1

\[\sum_{n=1}^{\infty}\frac{ 1 }{ (n+3)(n+4) }\]

Answer 2

Use partial fractions

Answer 3

\[\frac{1}{4}- \lim_{n \rightarrow \infty}\frac{ 1 }{ n+4 } =1 \]

Answer 4

@jiteshmeghwal9
i dont think you should do that, you get
\( \infty - \infty \)

Answer 5

If you start writing out terms you get
1/(1+3) - 1/(1 + 4)
+
1/(1+4) - 1/(1+5)
+ ...
you get a 'telescoping series'

Answer 6

whoops/ 1/4 is the answer

Answer 7

Don't know why I put 1 there.

Answer 8

sorry for that silly mistake @owen3

Answer 9

Is that correct?

Answer 10

it's an instructive mistake :)
$$\sum_{n=1}^\infty \frac{1}{n(n+1)} = \sum_{n=1}^\infty \left( \frac{1}{n} - \frac{1}{n+1} \right) \\
{} & {} = \lim_{N\to\infty} \sum_{n=1}^N \left( \frac{1}{n} - \frac{1}{n+1} \right) \\
{} & {} = \lim_{N\to\infty} \left\lbrack {\left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{N} - \frac{1}{N+1}\right) } \right\rbrack \\
{} & {} = \lim_{N\to\infty} \left\lbrack { 1 + \left( - \frac{1}{2} + \frac{1}{2}\right) + \left( - \frac{1}{3} + \frac{1}{3}\right) + \cdots + \left( - \frac{1}{N} + \frac{1}{N}\right) - \frac{1}{N+1} } \right\rbrack \\
= \lim_{N\to\infty} \left\lbrack { 1 - \frac{1}{N+1} } \right\rbrack = 1
$$

Answer 11

Woah.

Answer 12

was that your reasoning ?
@jojokiw3

Answer 13

Yep!

Answer 14

ran out of space. here it is again\[\sum_{n=1}^\infty \frac{1}{(n+3)(n+4)} = \sum_{n=1}^\infty \left( \frac{1}{n+3} - \frac{1}{n+4} \right) \\
= \lim_{N\to\infty} \sum_{n=1}^N \left( \frac{1}{n+3} - \frac{1}{n+4} \right) \\
= \lim_{N\to\infty} \left\lbrack {\left(\frac 1 4 - \frac{1}{5}\right) + \left(\frac{1}{5} - \frac{1}{6}\right) + \cdots + \left(\frac{1}{N+3} - \frac{1}{N+4}\right) } \right\rbrack \\
= \lim_{N\to\infty} \left\lbrack { \frac 1 4 + \left( - \frac{1}{5} + \frac{1}{5}\right) + \cdots + \left( - \frac{1}{N+3} + \frac{1}{N+3}\right) - \frac{1}{N+4} } \right\rbrack \\
= \lim_{N\to\infty} \left\lbrack { \frac 1 4 - \frac{1}{N+4} } \right\rbrack
\\ =\frac{1}{4}- \lim_{N \to \infty}\frac{ 1 }{ N+4 }
\\ =\frac 1 4 - 0
\]