Apply the Root Test to ∑n=1∞(n2n+1)n2\sum_{n=1}^{\infty} \left(\frac{n}{2n+1}\right)^{n^2}∑n=1∞(2n+1n)n2. Compute limn→∞ann\lim_{n \to \infty} \sqrt[n]{a_n}limn→∞nan where an=(n2n+1)n2a_n = \left(\frac{n}{2n+1}\right)^{n^2}an=(2n+1n)n2.
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