0
$\begingroup$

I'm reviewing questions from my calc textbook, and I've come across a couple of questions that I don't know how to solve. They are:

  1. $$\sum_1^\infty\frac{2^n+1}{3^n}$$
  2. $$\frac{\pi}6-\frac{\left(\dfrac{\pi}6\right)^3}{3!}+\frac{\left(\dfrac{\pi}6\right)^5}{5!}+\cdots$$

For the second problem, I was able to get the following, but I'm unsure where to go from here

$$\frac{\left((-1)^{n+1}\right)\left(\dfrac{\pi}6\right)^{2n-1}}{(2n-1)!}$$

$\endgroup$
  • $\begingroup$ The formate is important, as for example in this case it is very hard to understand (2), say. But it'd also be interesting, and even relevant, to know what've you tried to solve this, where does this come from, etc. $\endgroup$ – DonAntonio Oct 23 '16 at 22:48
  • $\begingroup$ Yes, achille, I did, thank you for pointing that out $\endgroup$ – M. Daroo Oct 23 '16 at 22:55
1
$\begingroup$

$$\sum_{n=1}^{\infty }\frac{2^n+1}{3^n}=\sum_{n=1}^{\infty }(\frac{2}{3})^n+\sum_{n=1}^{\infty }\frac{1}{3^n}$$ then use the geometric series which is $${\frac {1}{1-x}}=1+\sum _{n=1}^{\infty }x^{n}\quad {\text{ for }}|x|<1$$ or $${\frac {x}{1-x}}=\sum _{n=1}^{\infty }x^{n}$$

For the second series, you can depend on the following series $$\sin x=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{(2n+1)!}}x^{2n+1}=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-\cdots $$ at $x=\frac{\pi}{6}$

$\endgroup$
  • $\begingroup$ Nice answer +1 . Yet we'll still have to see whether the OP meant $\;2^n+1\;$ or $\;2^{n+1}\;$ in his first question. It doesn't matter, though, as the way is marked. $\endgroup$ – DonAntonio Oct 23 '16 at 23:01
  • $\begingroup$ @DonAntonio Thank you so much $\endgroup$ – E.H.E Oct 23 '16 at 23:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.