Evaluating a Limit with Cosine Functions

[sit.think.solve]

Evaluate the following limit:
$$\lim_{n\rightarrow \infty}\frac{1+\cos(\frac{x}{n})+...+\cos(\frac{n-1}{n}x)}{n}$$

 

[tiny-tim]

Hi sit.think.solve!

Hint: learn your trigonometric identities …

what is cosA + cosB? ​

 

[daudaudaudau]

You can write the sum on closed form if you think about cosine as the real part of the complex exponential.

 

[flatmaster]

There might be a problem here. The sum is discrete while the limit is continuous. Is there any practical application here or is this just math?

 

[sit.think.solve]

Thanks for the tips,

having some trouble trying to reduce the sums though:

Presumably I should add,

$$\cos(\frac{0}{n}x)+\cos(\frac{n-1}{n}x)$$
$$\cos(\frac{1}{n}x)+\cos(\frac{n-2}{n}x)$$

and so on, but then what would be the last term in such sum? In fact, can I do sums like this?

 

[daudaudaudau]

Like this
$$\sum_{k=0}^{n-1} \cos{\frac{kx}{n}}=\Re{\left(\sum_{k=0}^{n-1} \exp{i\frac{kx}{n}}\right)}$$

which is a geometric series. I don't think it converges to anything particular as n->infinity, but it is bounded when x is not equal to zero...

 

[tiny-tim]

Hi sit.think.solve!

… Presumably I should add,

$$\cos(\frac{0}{n}x)+\cos(\frac{n-1}{n}x)$$
$$\cos(\frac{1}{n}x)+\cos(\frac{n-2}{n}x)$$

and so on, but then what would be the last term in such sum? In fact, can I do sums like this?

Yes, of course you can …

and you should get a common factor which you can then put outside a bracket.

Alternatively, use daudaudaudau's method, and remember that each term inside the ∑ is (e^ix/n)^k