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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.

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    $\begingroup$ Ratio test will work: the ratio of successive numerators behaves like $7$ while the ratio of successive denominators behaves like $k$. $\endgroup$ – Ian Jun 1 '18 at 19:09
  • $\begingroup$ Please show us the work you've done in using the ratio test, and the ratio you set up for taking the limit of. Perhaps you made a mistake in your set up. Perhaps we can show you, then, how to proceed from where you left off. But please don't claim to have done anything at all unless you're willing to provide your work, in the body of your question. $\endgroup$ – Namaste Jun 1 '18 at 21:50
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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

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

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

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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.

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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.

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