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I've been at this problem for days.

Forgot most of my series from calculus, so I started to review series and sequence for numerical methods and approximations.

$$\sum_{n=5}^{\infty} \frac{12}{16n^2+40n+21}$$

How can I evaluate this series using telescoping method?

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Hint

$$16n^2+40n+21=(4n+3)(4n+7)$$

$$\frac{12}{16n^2+40n+21}=\frac{3}{4n+3}-\frac{3}{4n+7}=\frac{3}{4n+3}-\frac{3}{4(n+1)+3}$$

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  • $\begingroup$ hello, I tried this and got 5/8 which doesn't seem to be correct. perhaps I've done my telescoping incorrectly. $\endgroup$ – Calilaun Aug 15 '16 at 21:59
  • $\begingroup$ @Calilaun Since the series is telescopic, you are left with the first term which is $\frac{3}{4 \cdot 5+3}=\frac{3}{23}$. $\endgroup$ – N. S. Aug 15 '16 at 22:44
  • $\begingroup$ @Calilaun More technically, you are left with the first and last terms, but the last term is $0$. $\endgroup$ – Simply Beautiful Art Aug 15 '16 at 23:46
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Hint: $$\frac { 12 }{ 16n^{ 2 }+40n+21 } =\frac { 12 }{ 16n^{ 2 }+40n+25-4 } =\frac { 12 }{ { \left( 4n+5 \right) }^{ 2 }-4 } =\frac { 12 }{ \left( 4n+3 \right) \left( 4n+7 \right) } =3\left( \frac { 1 }{ 4n+3 } -\frac { 1 }{ 4n+7 } \right) $$

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Firstly, it's clear that this series converges as it looks roughly like $$ \sum_{n \geq 1} \frac{1}{n^2},$$ which is convergent. This can be made rigorous by using what many calculus textbooks refer to as limit comparison.

This series does telescope. The way to see this is to expand the summands into partial fractions. To do this, you factor $16n^2 + 40n + 21 = (4n + 3) (4n + 7)$ and then figure out how to write $$ \frac{12}{16n^2 + 40n + 21} = \frac{A}{4n + 3} + \frac{B}{4n+7}.$$ Then write out the first several terms of your series and see what does and doesn't survive.

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  • $\begingroup$ thank you. Much of my confusion came from the fact that I forgot how to do partial fractions. :/ $\endgroup$ – Calilaun Aug 15 '16 at 21:59

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