# Calculate this sum

1. Feb 8, 2012

### 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
$$\frac{1}{1+2^{-2n+1}}$$?

2. Feb 8, 2012

### 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.

3. Feb 8, 2012

### 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 $10^{-8}$