<s>[INST] Use the sum of the first ten terms to approximate the summation $\sum_1^{\infty} \frac{8n}{(n+1)3^n}$ Use the sum of the first ten terms estimate the summation and then to estimate the error
$\sum_1^{\infty} \frac{8n}{(n+1)3^n}$
Okay, so i added up the first ten terms and got approximately 1.101237.
The answer key says:
Now, $ \frac{8n}{(n+1)3^n} < \frac{8n}{n3^n} = \frac{8}{3^n}$, and so the error is $R_{10} = \sum_{11}^{\infty} \frac{7}{3^n} \cong 7(\frac{\frac{1}{3^{11}}}{1-\frac{1}{3}}) \cong .000059$
Can somebody explain what's going on here? Why are we allowed and how do we know to make the simplification  $\frac{8n}{(n+1)3^n} < \frac{8n}{n3^n} = \frac{8}{3^n}$? Will the infinite summations of these two things be the same since $n$ is going to infinity anyway?? It seemed a bit arbitrary. My textbook doesn't say anything about this. Thanks!!
 [/INST] This method leads to a conservative estimation for the error since
$$\frac{8n}{(n+1)3^n} < \frac{8n}{n3^n} = \frac{8}{3^n} \implies R_{10, exact} = \sum_{n=11}^{\infty} \frac{8n}{(n+1)3^n}< R_{10, approx} = 8\sum_{n=11}^{\infty} \frac{1}{3^n}$$
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