Question

72–86. Evaluating series Evaluate each series or state that it diverges.∑ (k = 2 to ∞) ln((k + 1)k⁻¹) / (ln k × ln(k + 1))

Accepted Answer

First, rewrite the general term of the series to simplify the expression inside the summation. The term is given by $$\frac{\ln\$$left((k+1)k^{-1}$$ight)}{\ln k 	imes \ln(k+1)}. $$Use the logarithm property $$\ln(a/b) = \ln a - \ln b to $$rewrite the numerator as $$\ln(k+1) - \ln k. $$Substitute the simplified numerator back into the term to get $$\frac{\ln(k+1) - \ln k}{\ln k 	imes \ln(k+1)}. $$Then, separate this fraction into two parts: $$\frac{\ln(k+1)}{\ln k 	imes \ln(k+1)} - \frac{\ln k}{\ln k 	imes \ln(k+1)}. $$Simplify each part of the separated fraction. The first part simplifies to $$\frac{1}{\ln k}$$ and the second part simplifies to $$\frac{1}{\ln(k+1)}. So $$the general term becomes $$\frac{1}{\ln k} - \frac{1}{\ln(k+1)}. $$Recognize that the series is telescoping because each term is of the form $$a_k - a_{k+1}$$, where $$a_k = \frac{1}{\ln k}. $$Write out the first few terms explicitly to see the cancellation pattern. Use the telescoping property to express the partial sum $$S_n = \sum_{k=2}^n \left( \frac{1}{\ln k} - \frac{1}{\ln(k+1)} ight) as$$ $$\frac{1}{\ln 2} - \frac{1}{\ln(n+1)}. $$Then analyze the limit of $$S_n as$$ $$n 	o \infty to $$determine whether the series converges or diverges.

72–86. Evaluating series Evaluate each series or state that it diverges.
∑ (k = 2 to ∞) ln((k + 1)k⁻¹) / (ln k × ln(k + 1))

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Key Concepts

Convergence and Divergence of Infinite Series

Properties of Logarithms

Comparison and Limit Comparison Tests