Question

How do you use the direct comparison test to determine if `Sigma 1/(n^2+1)` from `[1,oo)` is convergent or divergent?

Answer 1

If we have two sequences that fit the criteria `0 < a_n < b_n`, then we have two cases:

• If `sumb_n` is known to be convergent, then `suma_n` is convergent also.
• If `suma_n` is known to be divergent, then `sumb_n` is divergent also.

Notice that `1/(n^2+1)` is very similar to the function `1/n^2`. Consider which of the two is larger.

Note that `1/(n^2+1)` has a larger than denominator than `1/n^2`. Fractions with larger denominators are smaller (consider `1//8` versus `1//5` — the one with the larger denominator is smaller).

Thus, `1/(n^2+1)<1/n^2`.

On `n in [1,oo)`, we see that `0<1/(n^2+1)<1/n^2`.

From the p-series test, we know that `sum_(n=1)^oo1/n^2` is convergent.

Since `1/(n^2+1)<1/n^2`, we know that `sum_(n=1)^oo1/(n^2+1)` is convergent as well via the direct comparison test.