sum_(n=1)^(oo)ln(n)-ln(n+1)

Recognize Telescoping Series: Recognize that the series is telescoping. The terms ln(n) and -ln(n+1) will cancel out with the subsequent terms in the series, except for the first negative term.Write Out Terms: Write out the first few terms of the series to see the pattern. sum_(n=1)^(oo)(ln(n)-ln(n+1)) = (ln(1) - ln(2)) + (ln(2) - ln(3)) + (ln(3) - ln(4)) + ...Cancel Out Terms: Notice that all the terms ln(2) to ln(n) cancel out, and we are left with only the first and the last terms. The series simplifies to ln(1) - ln(n+1) as n approaches infinity.Evaluate Remaining Terms: Evaluate the remaining terms. ln(1) is 0 because the natural logarithm of 1 is 0. As n approaches infinity, ln(n+1) also approaches infinity.Determine Sum: Determine the sum of the series. Since ln(n+1) approaches infinity, the series does not converge to a finite sum. Therefore, the sum of the series is negative infinity.

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Algebra 2

Sum of finite series starts from 1

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\sum_{n=1}^{\infty}\ln(n)-\ln(n+1)

Recognize Telescoping Series: Recognize that the series is telescoping. The terms $\ln(n)$ and $-\ln(n+1)$ will cancel out with the subsequent terms in the series, except for the first negative term.
Write Out Terms: Write out the first few terms of the series to see the pattern. $\newline$ $\sum_{n=1}^{\infty}(\ln(n)-\ln(n+1)) = (\ln(1) - \ln(2)) + (\ln(2) - \ln(3)) + (\ln(3) - \ln(4)) + \ldots$
Cancel Out Terms: Notice that all the terms $\ln(2)$ to $\ln(n)$ cancel out, and we are left with only the first and the last terms. The series simplifies to $\ln(1) - \ln(n+1)$ as $n$ approaches infinity.
Evaluate Remaining Terms: Evaluate the remaining terms. $\ln(1)$ is $0$ because the natural logarithm of $1$ is $0$ . As $n$ approaches infinity, $\ln(n+1)$ also approaches infinity.
Determine Sum: Determine the sum of the series. $\newline$ Since $\ln(n+1)$ approaches infinity, the series does not converge to a finite sum. Therefore, the sum of the series is negative infinity.