2 test for convergence problems: $\sum^{\infty}_{k=1}5^k/(3^k+4^k)$and $\sum^{\infty}_{n=1}\tan\left(1/n\right)$

Question

For the first problem: $$\sum^{\infty}_{k=1}\frac{5^k}{3^k+4^k}$$

I tried to take the ratio test but was having trouble simplifying $$\frac{3^k + 4^k}{3^{k+1} + 4^{k+1}}$$

For the second problem: $$\sum^{\infty}_{n=1}\tan\left(\frac{1}{n}\right)$$

I don't know how to approach it.

Thanks for the help as always

lab bhattacharjee · Accepted Answer · 2013-03-30T09:56:20.267

4

For the first problem

$$\frac{\frac{5^{k+1}}{3^{k+1}+4^{k+1}}}{\frac{5^k}{3^k+4^k}}$$

$$=\frac{5^{k+1}(3^k+4^k)}{5^k(3^{k+1}+4^{k+1})}$$

$$=5\cdot\frac{\left(\frac34\right)^k+1}{3\cdot\left(\frac34\right)^k+4}$$

So, using Ratio Test: $$\lim_{k\to\infty}\frac{\frac{5^{k+1}}{3^{k+1}+4^{k+1}}}{\frac{5^k}{3^k+4^k}}=\frac54$$

edited Mar 30 '13 at 09:56

answered Mar 30 '13 at 09:51

lab bhattacharjee

274,582

How did you get the (3/4)^k ? – Billy Thompson Mar 30 '13 at 09:56
@BillyThompson, divide the numerator & the denominator by $4^k$ – lab bhattacharjee Mar 30 '13 at 09:57
1

Ok so I follow everything you did, but I don't know how you arrived at the final answer from the equation you gave directly above the final answer. Can you please clarify how that equates to 5/4? – Billy Thompson Mar 30 '13 at 10:09
2

@BillyThompson $\lim_{n\to\infty}x^n=0$ if $|x|<1$ right? – lab bhattacharjee Mar 30 '13 at 10:10
Ahh gotcha, thanks :) – Billy Thompson Mar 30 '13 at 10:14
@BillyThompson: see this also. http://math.stackexchange.com/a/327326/8581. – Mikasa Mar 30 '13 at 10:16

Mikasa · Answer 2 · 2013-03-30T19:12:57.757

3

Note that $$\lim_{n\to\infty}n^{+1}\tan(n^{-1})=1<\infty$$ so the corresponding series is divergent. That is similar to the following series: $$\sum_1^\infty\sin\left(\frac{1}n\right)$$

edited Mar 30 '13 at 19:12

answered Mar 30 '13 at 10:14

Mikasa

67,374

Ishan Banerjee · Answer 3 · 2013-03-30T10:00:03.460

2

For the 2nd one use the limit comparison test with $\dfrac 1 n$, to show it diverges. For the 1st one note that, $\dfrac{5^k}{3^k+4^k}>\frac 1 2(\dfrac{5}{4})^k$ which does not tend to $0$.

edited Mar 30 '13 at 10:00

answered Mar 30 '13 at 09:54

Ishan Banerjee

6,459

For the 2nd one how come you can compare to 1/n? it would be tan(1/n)/(1/n) right? and doesn't this go to 0 as the limit goes to infinity? I thought for LCT it has to reach a positive finite value. – Billy Thompson Mar 30 '13 at 10:01
It goes to 1. as $\dfrac{\tan x}{x}\to 1$ as $x\to0$ – Ishan Banerjee Mar 30 '13 at 10:04
is there some sort of rule in trig that tells you that x*tan(1/x) = tan(x)/x ? Because when I do LCT i get it in the form x.tan(1/x) ? – Billy Thompson Mar 30 '13 at 10:11
As $n\to \infty$ $\frac 1 n \to 0$ make the substitution and find the limit. – Ishan Banerjee Mar 30 '13 at 10:13
tan(0)/(1/n) right? – Billy Thompson Mar 30 '13 at 10:16
Let$ x= \dfrac{1}{n} $ then, $\dfrac{tan (\frac 1 n)}{\frac 1 n}=\dfrac{tan x}{x}$ and x tends to 0 as n gets bigger. – Ishan Banerjee Mar 30 '13 at 10:19
Figured it out, thanks – Billy Thompson Mar 30 '13 at 10:20

2 test for convergence problems: $\sum^{\infty}_{k=1}5^k/(3^k+4^k)$and $\sum^{\infty}_{n=1}\tan\left(1/n\right)$

3 Answers3