# Tips on intuition for solving these convergent/divergent series problems? [closed]

I have really bad intuition when figuring out what test/approach I should use for checking whether a series converges. This definitely stems for my weak foundation in math overall but specifically algebra/rearranging equations; I can't visualize quickly that one equation "looks like" another. Could you give me some tips on how you would approach these?

Example 1: I initially started this as a ratio test but when I simplified it I ended up getting divergent as the answer which is wrong... $$A_n=\frac{n}{n+\sqrt{n}}$$ Example 2: The question itself said to use the integral test, but I probably wouldn't have thought of it myself. Furthermore my answer ended up as $$\infty - \infty$$ and I wasn't sure what this would mean $$DNE\implies divergent$$ (doubt that's right in the first place). $$A_n=\sum^{\infty}_{x=1} \frac{ln(x)}{x}$$ Example 3: I thought this was the use of the addition rule where convergent + conv/divergent $$\implies convergent$$ but turns out the answer was divergent. Does the addition rule not apply for subtraction?? $$A_n=\sum^{\infty}_{x=1}\frac{1}{x}-\frac{1}{x^2}$$ If I'm allowed, I will edit this question with a few more examples so that I can really drill things home but idk if that's not allowed so holding back for now.

ALSO, to find what it converges to we've only used the limit as $$n \to \infty$$ of the $$S_n=\frac{A_0(1-r^k)}{1-r}$$formula, where we get $$r$$ by calculating a few sample values and dividing $$A_{n+1}/A_n$$ but that doesn't seem to be working as I get to questions with very obscure looking $$A_n$$. Is there another way (at the undergraduate Calc.3 level)?

• Are you sure that example 1 is not divergent? – gune Oct 23 '19 at 1:13
• According to the answers, it "converges to 1". I will ask around if there's been corrections but this is already an updated copy of the answer key... – Five9 Oct 23 '19 at 1:19
• Shouldn't that diverge because: $\frac{1}{2n}=\frac{1}{n+n}\leq\frac{n}{n+n}\leq\frac{n}{n+\sqrt{n}}$ and the $\frac{1}{2}\sum\frac{1}{n}$ diverges – gune Oct 23 '19 at 1:29
• $A_n = \frac{n}{n+\sqrt{n}} = \frac{n}{n}\frac{1}{1+\frac{1}{\sqrt{n}}} = \frac{\sqrt{n}}{\sqrt{n}+1}$ – Vaas Oct 23 '19 at 1:30
• Examples 2 and 3 are written incorrectly. The letter $n$ is used in two different ways. – littleO Oct 23 '19 at 3:21

1. Do the terms converge to $$0$$? E.g. $$\frac {n}{n+\sqrt n}=\frac {1}{1+1/\sqrt n}\to 1.$$

2. Are the terms larger (eventually) than the terms of a positive series that you know to be divergent? E.g. $$\frac {\ln n}{n}>\frac {1}{n}$$ when $$n\ge 3.$$ And if $$n\ge 4$$ then $$\frac {n}{n+\sqrt n}\ge \frac {n}{n+(n/2)}=2/3.$$

3. Are the terms comparable to $$\frac {p(n)}{q(n)}$$ for some polynomials $$p(n),q(n)$$? If $$p\ne 0$$ then $$\sum_n\frac {p(n)}{q(n)}$$ converges iff $$deg(q)\ge 2+deg(p).$$

4. If $$\sum_n B_n$$ converges absolutely then $$\sum_n(A_n+B_n)$$ converges iff $$\sum_n A_n$$ converges. E.g. $$A_n=\frac {1}{n}$$ and $$B_n=-\frac {1}{n^2}.$$ For this example we can also use 2., as $$\frac {1}{n}-\frac {1}{n^2}=\frac {p(n)}{q(n)}$$ with $$p(n)=n-1$$ and $$q(n)=n^2.$$

5. The Cauchy Condensation Test. If $$a_n\ge a_{n+1}\ge 0$$ for all but finitely many $$n$$ then $$\sum_na_n$$ converges iff $$\sum_n2^na_{2^n}$$ converges. E.g.$$\sum_{n\ge 2}\frac {1}{n(\ln n)^k}$$ converges iff $$\sum_{n\ge 2}\frac {1}{n^k(\ln 2)^k}$$ converges... and by a 2nd application of this test, $$\sum_{n\ge 2}\frac {1}{n^k}$$ converges iff $$\sum_{n\ge 2}\frac {2^n}{2^{nk}}=\sum_{n\ge 2}\frac {1}{2^{n(k-1)}}$$ converges iff $$k>1....$$ so $$\sum_{n\ge 2}\frac {1}{n(\ln n)^k}$$ converges iff $$k>1.$$

1. Alternating series. If $$a_n\ge a_{n+1}\ge 0$$ for all but finitely many $$n,$$ and if $$a_n\to 0$$ then $$\sum_n(-1)^n a_n$$ converges. E.g. $$a_n=1/n.$$
• 1. Isn't the limit test supposed to be divergent when the limit is zero? I think I'm mixing up concepts. AFAIK the limit test takes $lim_{n\to\infty}A_n = L$ and if $L \neq 0$, divergent if $L=0$, gives no information.  2. Your example is helpful but could you demonstrate via the integral test, because the question asked me to do an integral test and wasn't able to do it  – Five9 Oct 23 '19 at 5:12
• 3. I took your hint but not exactly as given, because I wasn't sure why it needed to be 2 + deg(p). I got the $\frac{n-1}{x^2}$ equation but tried to do a ratio test on it and failed. Could you re-explain? – Five9 Oct 23 '19 at 5:18
• If $a_{n+1}/a_n\to 1$ the ratio test is of no use. – DanielWainfleet Oct 23 '19 at 13:45