Find Sum of Series to Within 0.01

[ineedhelpnow]

how many term of the series $\sum_{n=2}^{\infty}\frac{1}{[n(ln (n))^2]}$ would you need to add to find its sum to within 0.01?

approximate the sum of the series correct to four decimal places.

$\sum_{n=1}^{\infty}\frac{(-1)^n}{3^nn!}$

 

[zzephod]

how many term of the series $\sum_{n=2}^{\infty}\frac{1}{[n(ln (n))^2]}$ would you need to add to find its sum to within 0.01?

Because the terms of the sum are positive and decreasing:

$$\sum_{n=N+1}^{\infty}\frac{1}{[n(ln (n))^2]}<\int_{N}^{\infty}\frac{1}{[x(ln (x))^2]}\;dx=\frac{1}{\ln(N)}$$

Thus the error in truncating the sum to $N$ terms is less than $1/\ln(N)$. So if you find any $N$ such that $1/\ln(N)\le 0.01$ the error in truncating the sum to $N$ terms will be less than $0.01$.

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[zzephod]

approximate the sum of the series correct to four decimal places.

$\sum_{n=1}^{\infty}\frac{(-1)^n}{3^nn!}$

Because this is an alternating series of decreasing terms the error in truncating the sum is less than the absolute value of the first neglected term.

$$\left|\sum_{n=1}^{\infty}\frac{(-1)^n}{3^nn!}-\sum_{n=1}^{N}\frac{(-1)^n}{3^nn!}\right|< \frac{1}{3^{N+1}{(N+1)}!}$$

So the error is less than $0.00005$ when $N\ge 4$

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[ineedhelpnow]

for the first one:

is this right? i followed another example i saw but i don't know if my work makes that much sense.

$\int_{x=n}^{x=\infty} \ \frac{dx}{x(ln(x))^2}<0.01$

$\frac{-1}{ln(x)}]_{n}^{\infty}<0.01$

$0-(-\frac{-1}{ln(n)})<0.01$

$ln(n)>100$

$n>e^{100}$

 

[ineedhelpnow]

i don't understand the second one at all. i don't need to find the error though.

 

[zzephod]

for the first one:

is this right? i followed another example i saw but i don't know if my work makes that much sense.

$\int_{x=n}^{x=\infty} \ \frac{dx}{x(ln(x))^2}<0.01$

$\frac{-1}{ln(x)}]_{n}^{\infty}<0.01$

$0-(-\frac{-1}{ln(n)})<0.01$

$ln(n)>100$

$n>e^{100}$

That is correct.

 

[zzephod]

i don't understand the second one at all. i don't need to find the error though.

You need to find the number of terms $N$ for which the result of the truncated sum is correct to four decimal places. Which requires that the error be less than $0.00005$

As stated in the previous post this occurs when:

$$\frac{1}{3^{N+1}(N+1)!} <0.00005$$

Which holds for any N\ge 4. Hence:

$$S=\sum_{n=1}^{\infty}\frac{-1}{3^{n} n!} \approx \sum_{n=1}^{4} \frac{-1}{3^{n}n!}$$

and the approximation is correct to four decimal places.

Except it turns out that we have to be a bit more careful because of rounding to that fourth decimal place, so let's look at truncation to 4,5 and 6 terms, define:

$$S(N)=\sum_{n=1}^{N}\frac{-1}{3^{n} n!}$$

Then $S(4)=-0.2834362139918$, $S(5)=-0.2834705075446$, $S(6)=-0.2834686023472$, so it looks like we actually need 5 terms in the sum to get it to round correctly to 4 decimal places..