Question

Summation {n=1)^infty sin(1/n)
Show that this diverges

Thanks!

Answer 1

using the fact that \(\sin x\approx x\) for small \(x\);
$$
\large
\sum_{n=1}^\infty \sin \cfrac{1}{n}\approx \sum_{n=1}^\infty \cfrac{1}{n}
$$
which diverges: http://en.wikipedia.org/wiki/Harmonic_series_%28mathematics%29

Answer 2

awesome, ok that's kinda what i thought, but the sin threw me off big time.

Answer 3

we treat sin as a constant that isnt sin(1/n) but (sin)(1/n)?

Answer 4

No, sin needs a parameter. The parameter is 1/n.

Answer 5

so don't you have to include the sin in the comp. series you are comparing the given series to?

Answer 6

that was worded terribly.. i mean shouldn't the sum of 'b_n' be sin(something less than 1/n)

Answer 7

Really what we are doing is saying that at some point forward \(\text{i.e }\exists~ n>N \) such that \(\sin x \approx x\). So, the part of the sum before this is a finite sum. So if you add a finite sum to another sum that diverges, in this case the harmonic series, the whole thing diverges.

Answer 8

The approximation of sin x to x is exact the smaller x is. so the say that one diverges less than the other does not make sense in this context.

Take a calculator and do sin .01. It will be about .01, do it for smaller numbers, you'll see the effect. Also, if you know about taylor series expansion, you can do this for sin x and the approximation will be just x for small x: http://www.wolframalpha.com/input/?i=taylor+series+sin%28x%29

Does that make sense?

Answer 9

this makes sense, when you say 'small x' do you mean, as x gets smaller and smaller \[\sin (x) \approx x\]
I'm look at the link you sent me of the Taylor series expansion.. what i see is that as the terms increase, sin(x) oscillates between -1 and 1.. so it doesn't converge..

and because x is essentially sin(x) as you go to infinity - we can conclude that sin(1/n) is only slightly greater than (almost equal to) 1/n? hopefully this makes sense!
@ybarrap

Answer 10

$$
\large
\sin x = x-x^3/6+x^5/120+O(x^7)
$$

Note that the terms \(x^3, x^5,\cdots\) are sooooo insignificant compared to x when x is small. That is why sin x = x when x is small. So sin(1/n) is small when n is large, that's saying the same thing as sin x is small when x is small.

That the larger order terms oscillate between + and - is not important because their MAGNITUDES are so insignificant when n is large (i.e. 1/n is small).

At large n sin 1/n = 1/n and is no longer approximate, it is exact from an analytical perspective -- they both reach the asymptote (sin 1/n = 1/n \(\to \) 0) so they are equal (i.e. they both diverge), those extra terms in the taylor series go off to zero in the limit.

Answer 11

One more thing about this to visualize. To say that sin x = x for small x is to say that the graph of the taylor series approaches x with small x around the origin (i.e. x=0,y=0). You can see this linear (i.e. y=sin x = x) behavior in the following graph around this point:

Answer 12

so basically between -1 and 1, x is the best approximation for sin(x). This is what the graph looks like.. also I just realized this is Calc 3 (multivariable calc), i'm in calc 2 so no wonder this seems new. But i see how this makes sense, the pattern seems obvious now (as does the need for taylor series approximations)! wohoo, math!

Answer 13

and thank you @ybarrap!

Answer 14

lol, yw!