Quick convergent series question.

1. Dec 30, 2012

binbagsss

2^n + n!/4^n ?
so by the 'vanishing condition' as n!/4^n does not ---> 0 as n --> infinity, this part of the series diverges.
however (e.g via the ratio test 2^n/4^n converges).
My book concludes that due to part a, the entire series divegres. However I am struggling to see how this justifies that the whole series will diverge - I thought that if part of a series diverges and part diverges , than what the whole series does depends upon the exact convergence and divergence - i.e could diverge or converge?

2. Dec 30, 2012

lurflurf

The whole series will diverge by comparison

n!/4^n<(2^n + n!)/4^n

If the parts were of opposite sign there would be the possibility of convergence.

3. Dec 30, 2012

Ray Vickson

Looking at what you wrote, which is
$$2^n + \frac{n!}{4^n},$$
we have, of course, that both parts diverge. However, assuming you meant
$$\frac{2^n + n!}{4^n},$$
which, for some reason you did not want to write as (2^n + n!)/4^n, then you are correct: one "part" converges and the other diverges. However, that *automatically* means that the total diverges. Suppose we have two parts are $t_1(n)$ and $t_2(n)$ and that$\sum t_1(n)$ converges while $\sum t_2(n)$ diverges. We have, for finite N:
$$S(N) = \sum_{n=1}^N (t_1(n) + t_2(n)) = S_1(N) + S_2(N),$$ where the $S_i(N)$ are the partial sums, and our assumptions are that
$$\lim_{N \to \infty} S_1(N) = s_1$$ exists and is finite, while
$$\lim_{N \to \infty} S_2(N)$$ does not exist (that is, it is either $\pm \infty$ or else does not exist at all, due to oscillations, etc). These two statements imply that S(N) does not have a finite limit as N → ∞, so the total series diverges.