Chapter 9: Problem 64
Determine the convergence or divergence of the series. \(\sum_{n=0}^{\infty} \frac{1}{4^{n}}\)
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Chapter 9: Problem 64
Determine the convergence or divergence of the series. \(\sum_{n=0}^{\infty} \frac{1}{4^{n}}\)
These are the key concepts you need to understand to accurately answer the question.
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$$ y=\sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2 n}}{(2 n) !} y^{n}+y=0 $$
show that the function represented by the power series is a solution of the differential equation. $$ y=1+\sum_{n=1}^{\infty} \frac{(-1)^{n} x^{4 n}}{2^{2 n} n ! \cdot 3 \cdot 7 \cdot 11 \cdot \cdots(4 n-1)}, y^{\prime \prime}+x^{2} y=0 $$
Let \(f(x)=\sum_{n=0}^{\infty} c_{n} x^{n}\), where \(c_{n+3}=c_{n}\) for \(n \geq 0 .\) (a) Find the interval of convergence of the series. (b) Find an explicit formula for \(f(x)\).
Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. The series \(\sum_{n=1}^{\infty} \frac{n}{1000(n+1)}\) diverges.
Find all values of \(x\) for which the series converges. For these values of \(x\), write the sum of the series as a function of \(x\). $$ \sum_{n=1}^{\infty} \frac{x^{n}}{2^{n}} $$
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