Use any method, determine whether the series $\sum_{k=1}^{\infty} \frac{7^{k} + k}{k! + 1}$ converges?

Question

This is the series $\sum_{k=1}^{\infty} \frac{7^{k} + k}{k! + 1}$. I began with the ratio test, but after I set up the ratio to take the limit of, I couldn't simplify the limit, since nothing would cancel out. I believe it converges, but I have no idea how to solve further after setting up that ratio test limit.

If someone has another method, that would be fine too. We are not allowed to use Comparison or Limit Comparison Tests.

Answer 1

One idea is to use a combination of the methods that you know.

We know that by the comparison test if the sum $$ \sum_{k = 1}^{\infty} \frac{7^k + k}{k!} $$ converges, then so does the sum that you are interested in. We can break this apart into two sums: $$ \sum_{k = 1}^{\infty} \frac{7^k}{k!} $$ and $$ \sum_{k = 1}^{\infty} \frac{k}{k!} = \sum_{k = 1}^{\infty} \frac{1}{(k - 1)!}. $$

These both converge by the ratio test, and hence so does the sum of these two sums.

Answer 2

Hint :$$\sum_{k=1}^{\infty} \frac{7^{k} + k}{k! + 1}\\\leq \sum_{k=0}^{\infty} \frac{7^{k} + k}{k! + 1}\\\sum_{k=1}^{\infty} \frac{7^{k} + 7^{k}}{k! + 1}\\\leq \sum_{k=1}^{\infty} \frac{2\times 7^{k} }{k! }=2\sum_{k=1}^{\infty} \frac{ 7^{k} }{k! }=2e^7$$

Answer 3

We have

$$\frac{7^{k+1}+k+1}{7^k+1} = \frac{7+\frac{k+1}{7^k}}{1+\frac1{7^k}} \xrightarrow{k\to\infty} 7$$

$$\frac{k!}{(k+1)!+1} = \frac{1}{k+\frac{1}{k!}} \xrightarrow{k\to\infty} 0$$

Therefore

$$\frac{a_{n+1}}{a_n} = \frac{7^{k+1}+k+1}{7^k+1} \cdot \frac{k!}{(k+1)!+1} \xrightarrow{k\to\infty} 0 $$

so your series converges.

Answer 4

Note that

$$\frac{7^{k} + k}{k! + 1}\sim\frac{7^{k}}{k!}$$

therefore the given series converges by limit comparison test with $\sum_{k = 1}^{\infty} \frac{7^{k}}{k!} $ which converges by ratio test

$$\frac{7^{k+1}}{(k+1)!}\frac{k!}{7^{k}}=\frac 7 {k+1}\to 0$$