Approximate the sum of each series to three decimal places.
$\sum {(-1)^n {1\over {n^3}}}$
From alternating series test, this series convergence.
$S\approx a_3+S_2$
$S\approx {1\over {27}} + {7\over {8}} \approx 0.912$
But WolframAlpha approximate the sum to $0.9015$ , what’s the wrong ?
find a positive integer $n$ such that $S_n$ approximates the sum of the series to 4 decimal places.
$\sum (-1)^n {1\over \sqrt{n}}$
How can I solve this?
Is this mean that the error will be less than $0.00001$ for example?
Best Answer
If an alternating series converges, then adding the first $n$ terms of the series will approximate the final (infinite) sum with error at most the next term in the series. Meaning if $S_n$ is the sum of the first $n$ terms, $S$ is the final infinite sum, and $a_{n+1}$ is the next term that you would add/subtract, then the error of approximating $S$ by $S_n$ is at most $|a_{n+1}|$, i.e. $|S-S_n|<|a_{n+1}|$.
In the first example, we want to find $n$ such that $$ |a_{n+1}|= \left|\dfrac{(-1)^{n+1}}{(n+1)^3} \right|= \dfrac{1}{(n+1)^3} <0.001= \dfrac{1}{1000}. $$ But then $(n+1)^3 > 1000$ so that $n+1 > \sqrt[3]{1000}=10$ and then $n>9$, so the first $10$ terms will do.
For the second, we want $$ |a_{n+1}| = \left| \dfrac{(-1)^{n+1}}{\sqrt{n+1}}\right| = \dfrac{1}{\sqrt{n+1}} < 0.0001= \dfrac{1}{10000}. $$ But then $\sqrt{n+1}> 10^4$ so that $n+1 > (10^4)^2=10^8=100,000,000$ then $n>99,999,999$ so you would have to add $100,000,000$ terms to get the first $4$ decimal digits.