How do you solve for the limit of this series? I need to take the limit of this summation so that I kind find out whether it converges or diverges. The equation is:

$$\sum_{k=1}^\infty \frac{4}{k+4}$$

What I have tried so far is the following:

$$\lim_{k \rightarrow \infty} \frac{k+4}{(k+1)+4}$$

Which then gives us $\frac{4}{4} = 1$, however I know that this is incorrect. How do your properly take the limit of the summation?
If you're just trying to find out weather it converges or diverges, you don't need to actually find its limit.

Instead try a comparison test with the Harmonic Series $$\sum_{n=1}^{\infty}\frac{1}{n}$$

This definitely diverges as it's a p-series with p=1
For $k\geq 1$,
$$
\frac{4}{4+k}=\frac{1}{1+\frac{k}{4}}&gt; \frac{1}{1+k}
$$

But, for $N\geq 1$
$$
\sum_{k=1}^{N}\frac{1}{1+k}=\frac{1}{2}+\frac{1}{3} + \ldots + \frac{1}{N+1}= \big(\sum_{k=1}^{N}\frac{1}{k}\big)+ \big(\frac{1}{N+1}-1\big)
$$

So, since the harmonic series diverges, given any $M&gt;0$, you can find an index $N$ such that $\sum_{1}^{N} 1/k &gt; M+1$, and since $1/(N+1)-1 &gt; -1$, for that same index N, $\sum_{1}^{N}1/(1+k)&gt;M$, and $\sum 1/(1+k)$ diverges.

Therefore, $\sum 4/(4+k)$ also diverges, by the comparison test.