What is the minimum number of terms you would need to sum in order
to approximate the sum of the series with an error
less than o.0001?

Ask by Mckenzie Parry. in the United States

Mar 26,2025

Question

What is the minimum number of terms you would need to sum in order to approximate the sum of the series $$ \sum_{n=1}^{\infty}(-1)^{n} \frac{4}{2 n^{\frac{6}{5}}+1} $$ with an error less than o.0001?

UpStudy AI · Accepted Answer

We can use the Alternating Series Estimation Theorem, which tells us that for an alternating series $$ \sum_{n=1}^{\infty} (-1)^n a_n, $$ if the sequence $$ a_n $$ is positive, decreasing, and tends to zero, then the error made by approximating the sum by the partial sum $$ S_N $$ is no larger than the absolute value of the next term: $$ \left| S - S_N \right| \leq a_{N+1}. $$ For the given series, the general term is $$ a_n = \frac{4}{2n^{\frac{6}{5}}+1}. $$ We require the error to be less than $$0.0001$$. Thus, we need $$ \frac{4}{2 (N+1)^{\frac{6}{5}}+1} < 0.0001. $$ ### Step 1. Solve the Inequality Begin with $$ \frac{4}{2 (N+1)^{\frac{6}{5}}+1} < 0.0001. $$ Multiply both sides by $$2 (N+1)^{\frac{6}{5}}+1$$ (which is positive for all $$N$$) and divide by $$0.0001$$: $$ 4 < 0.0001 \bigl(2 (N+1)^{\frac{6}{5}}+1\bigr). $$ This can be rewritten as $$ 2 (N+1)^{\frac{6}{5}}+1 > \frac{4}{0.0001}. $$ Since $$ \frac{4}{0.0001} = 40000, $$ the inequality becomes $$ 2 (N+1)^{\frac{6}{5}}+1 > 40000. $$ Subtract 1 from both sides: $$ 2 (N+1)^{\frac{6}{5}} > 39999. $$ Divide by 2: $$ (N+1)^{\frac{6}{5}} > 19999.5. $$ ### Step 2. Isolate $$ N+1 $$ Raise both sides to the power $$ \frac{5}{6} $$ (the reciprocal of $$ \frac{6}{5} $$): $$ N+1 > \left( 19999.5 \right)^{\frac{5}{6}}. $$ ### Step 3. Compute the Right‐Hand Side Approximately For estimation, notice that $$ 19999.5 $$ is very close to $$ 20000 $$. Thus, we approximate: $$ N+1 > (20000)^{\frac{5}{6}}. $$ Taking logarithms or using exponentials: $$ (20000)^{\frac{5}{6}} = e^{\frac{5}{6} \ln(20000)}. $$ We compute: $$ \ln(20000) = \ln(2 \times 10^4) = \ln(2) + \ln(10^4) \approx 0.6931 + 9.2103 = 9.9034. $$ Then, $$ \frac{5}{6} \ln(20000) \approx \frac{5}{6} \times 9.9034 \approx 8.2528. $$ Thus, $$ (20000)^{\frac{5}{6}} \approx e^{8.2528} \approx 3840. $$ ### Step 4. Determine the Minimum Number of Terms Since $$ N+1 $$ must be greater than approximately $$ 3840 $$, the smallest integer value satisfying this is $$ N+1 = 3841. $$ This means $$ N = 3841 - 1 = 3840. $$ ### Final Answer To ensure the error is less than $$0.0001$$, you must sum at least $$3840$$ terms. You need to sum at least 3840 terms to approximate the series with an error less than 0.0001.

What is the minimum number of terms you would need to sum in order
to approximate the sum of the series with an error
less than o.0001?

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What is the minimum number of terms you would need to sum in order to approximate the sum of the series with an error less than o.0001?