Dgdrxrt

the series in the problem is as follows, and we would like to see if the series converges, and what is the sum of infinite series:

$$sum_n=1^infty frac2^n+1+13^n$$

It's a homework problem, but I already asked my teacher about it at school and he didn't really help me that much.

The earlier problems that we had were easier than this one, because I cannot find the common ratio for this infinite series (infinite geometric series should have common ratio, and when you know it you can tell if there is convergence)

I read online after class, that what I should do is to find the limit of the sequence of partial sums.
This limit, if it exists and is a real-number, should be the value of the sum of the infinite series.

It makes sense conceptually when you have a large value of k, for a finite series $S_k$, when k becomes larger the finite sum becomes more and more like the sum of infinite series.

So that if I have partial sums in a sequence $S_1, S_2, S_3...S_k $
The larger the k, the better it is representing infinite series, assuming it converges at all, towards anything.

I managed to put some large values of n, into wolfram alpha and the correct answer seems to be that the infinite series converges and sum of infinite series is 4.5

But I don't have a solid idea how I could find the explicit formula for the sequence of partial sums for this particular infinite series. And I don't particularly have any other good ideas how to prove that the sum of infinite series should be 4.5 in this case...

$$sum_n=1^inftyfrac2^n+1+13^n=2sum_n=1^inftyleft(frac23right)^n+sum_n=1^inftyleft(frac13right)^n.$$Can you take it from here?

After 20 years, I still have to rederive the identity
$$ sum_n=1^infty alpha^n = fracalpha1-alpha tag1$$
whenever I need it. Since it isn't that difficult, perhaps we should do that first.

Suppose that $|alpha| < 1$, and let
$$ S_k := sum_n=1^k alpha^k = alpha + alpha^2 + dotsb + alpha^k $$
be the $k$-th partial sum. Then
$$ alpha S_k = sum_n=1^k alpha^k+1 = alpha^2 + alpha^3 + dotsb + alpha^k+1. $$
Subtracting, we obtain
beginalign
(1-alpha) S_k
&= S_k - alpha S_k \
&= left( alpha + alpha^2 + dotsb + alpha^k right) - left( alpha^2 + alpha^3 + dotsb + alpha^k+1 right) \
&= alpha + left( alpha^2 - alpha^2right) + left( alpha^3 - alpha^3right) + dotsb + left(alpha^k - alpha^kright) - alpha^k+1 \
&= alpha - alpha^k+1.
endalign
Solving for $S_k$, we get
$$ S_k = fracalpha - alpha^k+11-alpha. $$
Since $|alpha| < 1$, it follows that $lim_ktoinfty alpha^k+1 = 0$, and so
$$
sum_n=1^infty alpha^n
:= lim_kto infty sum_n=1^k alpha^n
= lim_ktoinfty fracalpha - alpha^k+11-alpha
= fracalpha1-alpha.
$$
But this is exactly the identity at (1). Huzzah!

Now, how do we use this here?

Observe that
$$ frac2^n+1 + 13^n
= frac2^n+13^n + frac13^n
= 2 left( frac23 right)^n + left( frac13 right)^n.
tag2$$
But series are linear, and so play nice with addition and multiplication. Specifically, if $sum a_n$ and $sum b_n$ converge, and $C$ is any constant, then we have
$$ sum_n=1^infty left[ a_n + b_n right] = sum_n=1^infty a_n + sum_n=1^infty b_n
qquadtextandqquad
sum_n=1^infty Ca_n = C sum_n=1^infty a_n. $$
These are basic rules for "simplifying" series, which are going to be quite useful for you. If you are familiar with integration, you can perform exactly these manipulations if you replace the $sum$ with $int$, and the sequences with functions. If you are not familiar with integration, ignore this comment for now, but maybe keep it in the back of your mind for later.
beginalign
sum_n=1^infty frac2^n+1 + 13^n
&= sum_n=1^infty left[ 2 left( frac23 right)^n + left( frac13 right)^n right] && (textby application of (2)) \
&= 2sum_n=1^infty left( frac23 right)^n + sum_n=1^infty left( frac13 right)^n && (textsimplify the series) \
&= 2 cdot fracfrac231-frac23 + fracfrac131- frac13 && (textby application of (1)) \
&= 2 cdot fracfrac231-frac23cdot frac33 + fracfrac131- frac13cdot frac33 && (textsilly, pedantic arithmetic) \
&= 2 cdot frac23-2 + frac13-1 && (textmore pedantic arithmetic) \
&= frac41 + frac12 && (textand some more) \
&= frac8+12 && (textalmost there) \
&= frac92. && (textdone!)
endalign

You can split the sum and use

$$sum_n=1^infty r^n=fracr1-r$$