Question

How would you solve this infinite series? `sum_(n=1) ^oo (5-cosn)/sqrtn`

Would you use the direct comparison test?

`sum_(n=1) ^oo (5-cosn)/sqrtn`

Answer 1

Diverges by the Direct Comparison Test.

Explanation:

The Direct Comparison Test seems like the best option. We do not want to end up with a limit involving a trig function, and with the Ratio Test or Limit Comparison Test, the cosine would not cancel out.

Furthermore, this does not appear to have an elementary indefinite integral, so the Integral Test is out of the question.

So, let's first recognize that `-1<cosn<1`. Not `<=, n ne pi/2+kpi` as it is an integer.

Then since we are never truly adding or subtracting a perfect `1` from the denominator, we can say with total assuredness that if

`a_n=(5-cosn)/sqrtn, b_n=(5-1)/sqrtn=4/sqrtn<=a_n` for all `n`.

Now, examine the series

`sum_(n=1)^oo4/sqrtn=4sum_(n=1)^oo1/n^(1/2)`

This is the `p`-series `sum1/n^p` with `p=1/2<1,` so the series must diverge.

Then, by the Direct Comparison Test, since the smaller series diverges, so does the larger series `sum_(n=1)^oo(5-cosn)/sqrtn`