Evaluate the sum S for the series: S = sum from n=1

Evaluate the sum $S$ for the series: $$ S = \sum_{n=1}^{\infty} \ln\left(1+\frac{1}{n}\right). $$

Step 1 · Decompose the general term

The general term is $a_n = \ln\left(1+\dfrac1n\right)$. Simplifying inside the logarithm:

$$ a_n = \ln\left(\frac{n+1}{n}\right). $$

Using $\ln(a/b) = \ln a - \ln b$:

$$ a_n = \ln(n+1) - \ln(n), $$

which is a difference $b_{n+1}-b_n$ with $b_n = \ln(n)$.

Step 2 · Determine the partial sum $S_N$

This is a telescoping series. The $N$-th partial sum is:

$$ S_N = \sum_{n=1}^{N} \big(\ln(n+1)-\ln(n)\big). $$

Writing out the terms:

$$ S_N = \big(\ln2-\ln1\big) + \big(\ln3-\ln2\big) + \cdots + \big(\ln N-\ln(N-1)\big) + \big(\ln(N+1)-\ln N\big). $$

The intermediate terms cancel, leaving the last positive term and the first negative term:

$$ S_N = \ln(N+1) - \ln(1) = \ln(N+1), $$

since $\ln(1)=0$.

Step 3 · Calculate the sum $S$

The sum $S$ is the limit of $S_N$ as $N \to \infty$:

$$ S = \lim_{N \to \infty} S_N = \lim_{N \to \infty} \ln(N+1). $$

Since $\ln(x) \to \infty$ as $x \to \infty$:

$$ \lim_{N \to \infty} \ln(N+1) = \infty. $$

The series diverges.

Summary Table

Additional Notes

• Even though the terms $a_n = \ln(1+\tfrac1n) \to 0$ as $n \to \infty$ (a necessary condition for convergence), this alone is not sufficient: the terms shrink too slowly, decaying like $\frac1n$ for large $n$, the same rate as the divergent harmonic series.
• Because $S_N = \ln(N+1)$ grows without bound (however slowly), this series is a standard example showing that $a_n \to 0$ does not guarantee convergence of $\sum a_n$.
• The divergence rate is extremely slow: to make $S_N$ exceed $10$, for instance, one would need $N+1 > e^{10}$, illustrating how logarithmic growth can mask divergence in numerical experiments.