Convergence of tan(1/n) using Direct Comparison Test

[whatlifeforme]

Homework Statement

Use any test to determine whether the series converges.

Homework Equations

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

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.

 

[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?

 

[STEMucator]

Homework Statement

Use any test to determine whether the series converges.

Homework Equations

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

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.

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

 

[LCKurtz]

Homework Statement

Use any test to determine whether the series converges.

Homework Equations

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

The Attempt at a Solution

Direct Comparison Test

tan(1/n) > 1/n

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

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

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

 

[whatlifeforme]

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

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

 

[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)

 

[LCKurtz]

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

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

 

[Mark44]

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?

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

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)

 

[whatlifeforme]

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

exactly.

 

[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.

 

[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?)

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

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.