$\sum_{n=1}^\infty a_n = 0.55 - (0.55)^3/3! + (0.55)^5/5! - (0.55)^7/7! + ...$
I am asked to find the no. of terms needed to approximate the partial sum to be within 0.0000001 from the convergent value of that series. Hence, I tried find the remainder, $R_n = S-S_n$ by the improper integral
$\int_{n}^{\infty} \frac{ (-1)^n(0.55)^{2n-1}} {(2n-1)!} dn $
However, I don't know how to integrate this improper integral so is there other method to solve this problem?
On
Your general sum, $$ \sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!} $$ is the power series of a trigonometric function. Note that I've corrected the limits a little bit so that the correct terms come out negative. Do you know what that function is? Given that, can you evaluate the sum?
On
Hint:
$\sin x=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+ \dots$
Your expression is simply $\sin (0.55)$.
On
Since this is an alternating series, the "alternating series estimation theorem" states that the "remainder" $R_n = | s - s_n | $ is no larger than the absolute value of the next term in the series $| a_{n+1} |$. The desired precision ($10^{-7}$ , in your case) can be set as greater than this term.
The way your summation is written, the general term is
$$ \ a_n = \frac{(-1)^{n-1} \cdot x^{2n-1}}{(2n-1)!} \ ,$$
so the inequality based on the "next term" will be
$$| \frac{(-1)^{(n+1)-1} \cdot x^{2(n+1)-1}}{(2[n+1]-1)!} | \ \le 10^{-7} \ \Rightarrow \ \frac{x^{2n+1}}{(2n+1)!} \ \le 10^{-7} \ .$$
This is not too convenient to think about, so we can take reciprocals in the inequality (remembering, of course, to reverse the sign!):
$$ \ \Rightarrow \ \frac{(2n+1)!}{x^{2n+1}} \ \ge \ 10^7 \ .$$
We are evaluating the series at $x = 0.55$, but it will not introduce any significant disturbance in the result to solve this for $ \ x = 0.5 \ $:
$$ \frac{(2n+1)!}{(\frac{1}{2})^{2n+1}} \ = \ 2^{2n+1} \cdot (2n+1)!\ \ge \ 10^7 \ .$$
Now, an inequality with a factorial in it is not going to be something we can solve "algebraically", but we can get an estimate for $ \ n \ $ pretty quickly if we know some factorial values. Since $ \ 2n + 1 \ $ is odd, we note that $ \ 9! \approx 360,000 \ $ and $ \ 11! \approx 40,000,000 \ $ , so starting from $ \ n = 4 \ $ would be a good first attempt:
$$2^{2 \cdot 4 \ + \ 1} \cdot (2 \cdot 4 \ + \ 1)! \ = \ 2^9 \cdot 9! \ \approx \ 500 \cdot 360,000 \ \approx \ 1.8·10^8 \ \ge \ 10^7 \ , $$
indicating that $ \ n = 4 \ $ certainly does the job. The only remaining question is whether we could cut the finite series back by one term more. Setting $ \ n = 3 \ $ yields
$$2^{2 \cdot 3 \ + \ 1} \cdot (2 \cdot 3 \ + \ 1)! \ = \ 2^7 \cdot 7! \ \approx \ 130 \cdot 5000 \ \approx \ 660,000 \ \le \ 10^7 \ , $$
so we cannot spare the $ \ n = 4 \ $ term. Your summation has an initial index of $ n = 1 $ , but the exponent in the general term is $ \ 2n - 1 \ $ , so we have need of five terms to reach the desired level of precision. (Indeed, we find that the fifth term produces $ \ + \frac{0.55^9}{9!} \ \approx \ 1.27·10^{-8} \ $ , so we have attained entirely adequate precision; the fourth term is only $ \ - \frac{0.55^7}{7!} \ \approx \ -3.0·10^{-6} \ $, which does not affect the seventh decimal place.)
[Warning to students: a "popular" (with exam-writers -- not necessarily exam-takers) multiple-choice question is to ask how many terms are needed for a given level of precision for a series. Always check what the initial index of the series is and which terms in the series have non-zero coefficients. It may be that a series has all the terms up to $ \ n = 8 \ $, but the count begins at $ \ n = 0 \ $, so in fact nine terms are required. In the problem here, the series runs to the ninth-power term, and the initial index is $ \ n = 1 \ $, but only five terms are used. So it is important to have an awareness of the specificiations for the series being evaluated.]
For an alternating series the error is less than the first neglected term, so just keep calculating until the terms get smaller than $10^{-7}$