# Find sum of series $\sum _{ n=1 }^{ \infty }{ { (-1) }^{ n+1 }\left( \frac { n+2 }{ { n }^{ 5 } } \right) }$ correct to 3 decimal places.

This is a question I came across while studying on Khan Academy:

Find the sum of the series $\sum _{ n=1 }^{ \infty }{ { (-1) }^{ n+1 }\left( \frac { n+2 }{ { n }^{ 5 } } \right) }$ correct to three decimal places.

I solved this by calculating ${ S }_{ 1 },{ S }_{ 2 },{ S }_{ 3 },...\\$ until I had a close enough value (bad method, I know). Anyway, here is the solution steps given in the answer:

Note that this is a convergent alternating series. The Alternating Series Estimation Theorem states that the error bound in the sum of the first $n$ terms is the absolute value of the first omitted term; i.e., $\left| { a }_{ n+1 } \right|$. To obtain three-digit accuracy, we need to find $n$ such that $\left| { a }_{ n+1 } \right| <\ 0.0005$, or equivalently $\frac { n+3 }{ { (n+1) }^{ 5 } } <0.0005$.

When $n=6$, we have $\left| { a }_{ n+1 } \right| =\frac { n+3 }{ { (n+1) }^{ 5 } } =\frac { 9 }{ 16807 } \approx .000535$; that is not accurate enough.

When $n=7$, we have $\left| { a }_{ n+1 } \right| =\frac { n+3 }{ { (n+1) }^{ 5 } } =\frac { 10 }{ 32768 } \approx .000305$; that will give a final result within $0.0005$.

Thus we need to find the sum of the first seven terms of the series.

$$\sum _{ n=1 }^{ 7 }{ { (-1) }^{ n+1 }\left( \frac { n+2 }{ { n }^{ 5 } } \right) \approx 2.8914 }$$

Hence, $2.891$ is accurate to three decimal places.

What I don't understand about this answer is if the estimate is $2.8914$ and the error bound is $0.000305$, I think the actual sum can be close to $2.8914+0.000305\approx 2.891705\approx 2.892$. Then the result above is not correct.

Does this answer take into account the fact that the $(7+1)th$ term is negative? Or is there something about rounding and error bounds that I don't understand?

Thank you.

• Indeed, 2.8914 means between 2.89135 and 2.89145 and 0.000305 is less than 0.0004 hence the sum is between 2.89135-0.0004 and 2.89145, which implies that 2.891 has three correct decimal places. (The same estimates for n=6 show it was not necessary to go to n=7.) – Did Oct 21 '15 at 7:27
• You are right, it is the fact that $\sum_1^7$ is an overestimate that enables us to conclude that $2.891$ is correct to 3 decimal places. – André Nicolas Oct 21 '15 at 7:28
• To the op: If you want it back I'll undelete my answer, but it seems unnecessary given you're already pretty much there yourself. – Adam Hughes Oct 21 '15 at 7:32
• @Did Thank you. But do you mean if somehow the estimate is an underestimate, the sum can be between 2.89135 and 2.89145+0.0004. Then 2.892 would be correct, right? – user52139 Oct 21 '15 at 7:49
• @AdamHughes Thank you very much. I actually need it back if you don't mind, please. – user52139 Oct 21 '15 at 7:52

I would read the question the same way you did-find a three decimal value that you know the true sum of the series rounds to. We know from the alternating series theorem that the truncation error is smaller than the first neglected term in absolute value and of the same sign. There is no truncation error that we can guarantee will be small enough, because we could be poised right on the rounding boundary, with the computed answer something like $2.8915$. Here, because after seven terms we have $2.8914$ and we know the eighth term is negative, the correct answer is in the range $(2.8911,2.8914)$ and this entire range rounds to $2.891$

• +1, warning about problem with rounding boundary is well placed here. This answer might be further improved by pointing out the problem of numerically finding any decimal of $\sum_{n=0}^\infty \frac 1{2^n}$. – Ennar Oct 21 '15 at 10:38

This is an alternating series, so the error is estimated by the last term. Since you want the error to be less than $10^{-3}$ in size, you need only choose $n$ so that ${n+2\over n^5}<10^{-3}$ since the terms are decreasing.

Solving we see $n+2<10^{-3}n^5$. We claim this is so for $n\ge 7$ as

$$7+2=9<16.807=7^5\cdot 10^{-3}.$$

For $n>7$ we see that by induction we may conclude this with the inductive step being

$$(n+1)+2<10^{-3}n^5+1< 10^{-3}(n+1)^5,$$

Then adding up the first $6$ terms gives you $2.891$ to three decimal places.