Direct Comparison Test

1. Apr 4, 2013

whatlifeforme

1. The problem statement, all variables and given/known data
Use any test to determine whether the series converges.

2. Relevant equations
$\displaystyle \sum^{∞}_{n=1} tan(1/n)$

3. The attempt at a solution
Direct Comparison Test

tan(1/n) > 1/n

By integral test: 1/n diverges thus, by dct, tan(1/n) diverges.

2. Apr 4, 2013

BruceW

looks good to me. They might want some more detail in your working, i.e. what is the range of values that 1/n can take. And why is tan(1/n) > 1/n for these range of values?

3. Apr 4, 2013

Zondrina

Yes, comparison test is what you want here. You may want to note that you need sufficiently large n for it to hold.

4. Apr 4, 2013

LCKurtz

As others have noted, you need to say why this is true.

It isn't 1/n that diverges, it is $\sum 1/n$ that diverges. Same for tan(1/n).

5. Apr 4, 2013

whatlifeforme

i thought it held only for sufficiently small n. it holds starting at n=1

6. Apr 4, 2013

BruceW

I think you are right that it holds, starting at n=1. But think about what would happen if you started at smaller n, for example starting at n=0.0001, would it still hold then?

Edit: and think about if it started at n=1000, would it hold? This should suggest whether it holds for sufficiently small or sufficiently large n.

Edit again: hint: look at the graph of tan(x), where might there be problems for the equation tan(x)>x ? And then think about how this applies to n (which is 1/x)

Last edited: Apr 4, 2013
7. Apr 4, 2013

LCKurtz

Think about the graphs of y = tan(x) and y = x for nonnegative x.

8. Apr 4, 2013

Staff: Mentor

Why would you want to check n = 0.0001? n is an index on a summation, typically limited to integer values.

9. Apr 4, 2013

whatlifeforme

exactly.

10. Apr 4, 2013

Infrared

It it doesn't seem that hard to show that $\tan \frac{1}{n} > \frac{1}{n}$ for all positive n. Since $\frac{1}{n} \leq 1$, it suffices to show that $\tan x \geq x$ for $0 \leq x \leq 1$ . Just let $f(x) = \tan x - x$ and find $f'(x) = \sec^2 x -1 \geq 0$. Since $f$ is zero at x=0 and increasing, it must be positive on $[0,1]$ as it is continous. If $\tan x - x$ is positive, then $\tan x \geq x$ and we are done.

11. Apr 4, 2013

BruceW

@HS-Scientist: yeah, don't give the game away though... (p.s. is this a British expression, or does this make sense to Americans, too?)

It doesn't have to be integer though. The point is that whatlifeforme was saying that n needs to be sufficiently small, but actually it holds for n sufficiently large. Although I admit, checking particular values of n is not very helpful. It is better to look at the graph.