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Estimate the value of the following convergent series with an absolute error less than 10^-4

∞
∑  ((-1)^k)/(2k+1)!
k=1

I set my ak=1/(2k+1)!, then sub in n+1 for k because I am approximating the value to 10^-4. However, I don't know how to handle the factorial. I am just left with (2n+3)!=10^4 and don't know how to solve for just n...

Verified Answer :

LammettHash

We can find the number of terms needed to fall within that accuracy. Taking the absolute difference of the \(n\)th and \((n+1)\)th partial sums, you have

\(\displaystyle\left|\sum_{k=1}^{n+1}\frac{(-1)^k}{(2k+1)!}-\sum_{k=1}^n\frac{(-1)^k}{(2k+1)!}\right|=\frac1{(2n+3)!}<10^{-4}=\dfrac1{10000}\)

or equivalently,

\((2n+3)!>10000\)

It's easy enough to check that \((2n+2)!\) exceeds 10000 for all \(n\ge3\), since \((2\times3+2)!=8!=362880\). This means the sum can be approximated to within \(10^{-4}\) with just the first three terms.

You have

\(\displaystyle\sum_{k=1}^3\frac{(-1)^k}{(2k+1)!}=-\frac1{3!}+\frac1{5!}-\frac1{7!}=-\frac1{7!}\left(\frac{7!}{3!}-\frac{7!}{5!}+1\right)=-\frac{799}{5040}\approx-0.158532\)

The actual value of the sum is \(\sin(1)-1\), which comes from the fact that

\(\sin x=\displaystyle\sum_{k=0}^\infty\frac{(-1)^kx^{2k+1}}{(2k+1)!}\implies \sin x-x=\sum_{k=1}^\infty\frac{(-1)^kx^{2k+1}}{(2k+1)!}\)

and plugging in \(x=1\). The actual value is approximately \(-0.158529\) which verifies that the approximation is within 4 decimal places of the infinite series.

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