Convergence of Series - a few things I don't understand

**OneMileCrash** · November 15th 2011, 08:03 AM

If the terms in a series approaches 0 as n grows without bound, how can the series possibly diverge? I don't understand that. I can only assume that the inverse is definitely true though - if the limit does not approach 0 then the series can never converge.

I've been solving the problems very awkwardly and usually incorrectly.

For example:

Infinite series:
n! / 6^n

My logic is that it must diverge simply because for very large n, n! becomes greater than 6^n, even though that's not the case at lower n.

For the infinite sum (2 + (3/n))^n, I just logically concluded that 2 does not matter at large n, than 3/n approaches 0, and (3/n)^n just approaches 0 faster, so in my mind the series has to converge. Something tells me that is not the case, though.

I believe my that my thinking is fine for the first, but wrong for the second. Can it still diverge if terms approach 0 as n grows without bound? An example and why, would be great!

Thank you!

**HallsofIvy** · November 15th 2011, 09:28 AM

Originally Posted by OneMileCrash

If the terms in a series approaches 0 as n grows without bound, how can the series possibly diverge? I don't understand that. I can only assume that the inverse is definitely true though - if the limit does not approach 0 then the series can never converge.

Yes, if the limit of the sequence of terms is not 0 the series does not converge. In order for the series to converge, the sequence of terms must converge to 0 fast enough. For example, the terms $\frac{1}{n}$ converge to 0 but the series $\sum_{n=1}^\infty \frac{1}{n}$ does NOT converge, by the "integral test". In a sense, that is the "critical example"- the series $\sum_{n=1}^\infty \frac{1}{n^p}$ converges if p> 1, diverges if $p\le 1$ .

I've been solving the problems very awkwardly and usually incorrectly.

For example:

Infinite series:
n! / 6^n

My logic is that it must diverge simply because for very large n, n! becomes greater than 6^n, even though that's not the case at lower n.

That's not bad but simpler is to use the "ratio test". If $a_n= n!/6^n$ then $a_{n+1}= (n+1)/6^{n+1}$ so that $\frac{a_{n+1}}{a_n}= \frac{6^{n+1}}{(n+1)!}\frac{n!}{6^n}= \frac{6}{n+1}$ which goes to 0 as n goes to infinity. A (non-negative) series will converge if and only if the sequence [itex]a_{n+1}/a_n[/tex] converges to a number less than 1.

[quote]For the infinite sum (2 + (3/n))^n, I just logically concluded that 2 does not matter at large n, than 3/n approaches 0, and (3/n)^n just approaches 0 faster, so in my mind the series has to converge. Something tells me that is not the case, though.[quote]
That 2 certainly does matter! Yes, 3/n approaches 0 which mean the sum basically involves adding 2 to itself an infinite number of times. Since you had already said that you understand that if the sequence of terms of a sum does not go to 0, the sum itself does not converge. Ok, $(2+ 3/n)^n$ does NOT go to 0 as n goes to infinity.

I believe my that my thinking is fine for the first, but wrong for the second. Can it still diverge if terms approach 0 as n grows without bound? An example and why, would be great!

Thank you!

Yes, a sum can still diverge if the terms approach 0. As I said before, they have to go to 0 "fast enough"- faster then 1/n. However, your thinking that $(2+ 3/n)^n$ goes to 0 is completely wrong! For large n, 3/n goes to 0 so this is close to [tex]2^n[/itex] which goes to infinity, not 0!

**OneMileCrash** · November 15th 2011, 09:36 AM

What was I thinking! Yes, that makes sense! It's more of the other way around, heh, the 2 is what matters the most. So what I should have done is look at what 3/n is doing, but at the same time remember that this is being added to 2. I suppose the only way convergence would be possible is if the n term approached -2, or something.

One last question. My professor told me very nearly the same thing. 1/n^2 converges because it is approaching 0 faster than 1/n. However, if we're dealing with an infinite number of terms, why does it matter how fast it approaches 0? I'm not sure I get that, because to me no matter how slow it is at approaching 0, an infinite number of terms seems to make that irrelevant in the grand scheme of its behavior. So what am I missing about that concept?

Furthermore, is 1/n some kind of defining standard (it sounds that way) for how fast the terms approach 0? If the terms approach 0 quicker than 1/n at all, convergence? Or is it not that black & white?

Thanks!

**emakarov** · November 15th 2011, 09:49 AM

Since I already typed this...

Originally Posted by OneMileCrash

If the terms in a series approaches 0 as n grows without bound, how can the series possibly diverge?

It turns out that when the terms are positive and approach 0 slowly enough, then the series can diverge. For example, $\sum_{n=1}^\infty 1/n$ diverges even though $\lim_{n\to\infty}1/n=0$ . In fact, $\sum_{n=1}^\infty 1/n$ grows as $\ln n$ : we have $\lim_{n \rightarrow \infty } \left( \sum_{k=1}^n \frac{1}{k} - \ln(n) \right)=\gamma\approx 0.577$ (Euler–Mascheroni constant).

The integral test says that $\sum_{n=1}^\infty f(n)$ converges iff $\int_1^\infty f(x)\,dx$ converges where $f:[1,\infty)\to\R_+$ is a positive and monotone decreasing function. We have $\int \frac{1}{x^a}\,dx=\frac{x^{1-a}}{1-a}+C$ if $a\ne 1$ and $\ln x+C$ if $a=1$ . Since $\lim_{x\to\infty}\ln x=\infty$ and $\lim_{x\to\infty}x^{1-a}<\infty$ iff $a>1$ , $\sum_{n=1}^\infty \frac{1}{n^a}$ converges iff $a>1$ .

On the other hand, when terms have alternating signs: $\sum_{n=1}^\infty (-1)^na_n$ where $a_n\ge 0$ , the series indeed converges if $a_n$ is monotone decreasing and $\lim_{n\to\infty}a_n=0$ regardless of how slowly $a_n$ tends to 0.

Originally Posted by OneMileCrash

I can only assume that the inverse is definitely true though - if the limit does not approach 0 then the series can never converge.

That's correct.

Originally Posted by OneMileCrash

Infinite series:
n! / 6^n

My logic is that it must diverge simply because for very large n, n! becomes greater than 6^n, even though that's not the case at lower n.

That's right.

Originally Posted by OneMileCrash

For the infinite sum (2 + (3/n))^n, I just logically concluded that 2 does not matter at large n, than 3/n approaches 0, and (3/n)^n just approaches 0 faster, so in my mind the series has to converge.

In fact, $\left(2 + \frac{3}{n}\right)^n>2^n$ , so the series diverges. You should also know the following remarkable limit:

$\lim_{n\to\infty}\left(1+\frac{a}{n}\right)^n=e^a$

So, $\left(2 + \frac{3}{n}\right)^n=2^n\left(1 + \frac{3/2}{n}\right)^n$ behaves like $2^ne^{3/2}$ for large $n$ .

Originally Posted by HallsofIvy

That's not bad but simpler is to use the "ratio test". If $a_n= n!/6^n$ then $a_{n+1}= (n+1)/6^{n+1}$ so that $\frac{a_{n+1}}{a_n}= \frac{6^{n+1}}{(n+1)!}\frac{n!}{6^n}= \frac{6}{n+1}$ which goes to 0 as n goes to infinity. A (non-negative) series will converge if and only if the sequence $a_{n+1}/a_n$ converges to a number less than 1.

In this case, $\frac{a_{n+1}}{a_n}=\frac{n+1}{6}\to\infty$ as $n\to\infty$ .

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