examples for limit comparison test

Example 1.

Does the following series converge?

$$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2+n+1}$$

The series is similar to ${\sum }_{n=1}^{\mathrm{\infty }}1/n^2$ which converges (use $p$-test, for example). Next we compute the limit:

$$\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{1}{n^2+n+1}}{\frac{1}{n^2}}=\underset{n\to \mathrm{\infty }}{lim}\frac{n^2}{n^2+n+1}=1$$

Therefore, since $1\ne 0$, by the Limit Comparison Test (with $a_n=1/(n^2+n+1)$ and $b_n=1/n^2$), the series converges.

Example 2.

Does the following series converge?

$$\sum _{n=1}^{\mathrm{\infty }}\frac{n^3+n+1}{n^4+n+1}$$

If we “forget” about the lower order terms of $n$:

$$\frac{n^3+n+1}{n^4+n+1}\sim \frac{n^3}{n^4}=\frac{1}{n}$$

and ${\sum }_{n=1}^{\mathrm{\infty }}1/n$ is the harmonic series which diverges (by the $p$-test). Thus, we take $b_n=1/n$ and compute:

$$\underset{n\to \mathrm{\infty }}{lim}\frac{\frac{n^3+n+1}{n^4+n+1}}{\frac{1}{n}}=\underset{n\to \mathrm{\infty }}{lim}\frac{n(n^3+n+1)}{n^4+n+1}=\underset{n\to \mathrm{\infty }}{lim}\frac{n^4+n^2+n}{n^4+n+1}=\underset{n\to \mathrm{\infty }}{lim}\frac{1+1/n^2+1/n^3}{1+1/n^3+1/n^4}=1$$

Therefore the series diverges like the harmonic does.