Calculate Sum Series: 3^n+1/4^n+2

[blob84]

Hello, I want to calculate the sum of this series:
\sum_{n=1}^\infty \frac{3^n+1}{4^n+2},
I know the series converges, I only found this useful way to factor the denominator:
4^n+2=2^{2n}+2=4^n(1+2^{-2n+1}),
now i have: \frac{3^n+1}{4^n+2}=(\frac{3^n}{4^n}+\frac{1}{4^n})(\frac{1}{1+2^{-2n+1}}),
i can calculate \sum_{n=1}^\infty \frac{3^n}{4^n}+ \sum_{n=1}^\infty \frac{1}{4^n}but what should be done with
\frac{1}{1+2^{-2n+1}}?

 

[kai_sikorski]

I'm not sure you can do the series you asked about, but you can get an upper bound if that helps by noticing that

\frac{1}{1-\frac{3 x}{4}}+\frac{1}{1-\frac{x}{4}}

is a generating function for

\frac{3^n+1}{4^n}

Maybe you could find a lower bound too and sandwich it.

 

[kai_sikorski]

By the way, this bound can be made really tight. For example, numerically we get

\sum _{n=1}^{\infty } \frac{1+3^n}{2+4^n} = 2.9142434104

Now since the majority of the difference between the series I proposed as an upper bound and the one you care about happens in the lower terms, evaluate a few of the lower terms exactly and do the remaining terms using the approximation.

\sum _{n=1}^{10 } \frac{1+3^n}{2+4^n} + \left(\frac{1}{1-\frac{3 }{4}}+\frac{1}{1-\frac{1}{4}} - \sum _{n=0}^{10 } \frac{1+3^n}{4^n} \right) = \frac{1747777419267069422319725824985}{599736246509717885804751618048}
= 2.914243435

So the difference is on the order of 10^{-8}