Based on the general term a_n, conclude that
Based on the general term $a_n$, conclude that the following series diverges: $$ \sum_{n=1}^{\infty} \frac{n}{n+1}. $$
Explanation of Method
The Divergence Test (the contrapositive of "convergent series have terms tending to zero") says that if $\lim_{n\to\infty} a_n \neq 0$, the series $\sum a_n$ cannot converge. Here the general term $\frac{n}{n+1}$ approaches $1$, not $0$, as $n\to\infty$, so the test immediately rules out convergence without needing any deeper analysis.
- Compute $\lim_{n\to\infty} a_n$.
- Compare the result to zero.
- Apply the Divergence Test to conclude.
Step 1 · Identify the general term
$$ a_n = \frac{n}{n+1}. $$
Step 2 · Calculate the limit of the general term
Dividing numerator and denominator by $n$:
$$ \lim_{n\to\infty} \frac{n}{n+1} = \lim_{n\to\infty} \frac{1}{1+\frac1n} = \frac{1}{1+0} = 1. $$
So $\lim_{n\to\infty} a_n = 1 \neq 0$.
Step 3 · Apply the divergence test
Since the general term does not tend to zero, the series $\sum_{n=1}^{\infty} \frac{n}{n+1}$ diverges. Intuitively, the terms $\frac12,\frac23,\frac34,\ldots$ all sit close to $1$, so the partial sums grow roughly like $n$ — essentially adding $1$ infinitely many times.
Summary Table
| Quantity | Value |
|---|---|
| General term $a_n$ | $\dfrac{n}{n+1}$ |
| $\lim_{n\to\infty} a_n$ | $1$ |
| Conclusion | Diverges (Divergence Test) |
Additional Notes
- The Divergence Test is a necessary but not sufficient condition: it can only prove divergence (when the limit is nonzero) — it can never prove convergence, even when $a_n\to0$, as the harmonic series famously shows.
- This series is a standard first example because the limit is easy to compute directly, without needing comparison tests, ratio tests, or integral tests.
- More generally, any rational function $a_n = \frac{p(n)}{q(n)}$ whose numerator and denominator have the same degree will have a nonzero limit (the ratio of leading coefficients), so $\sum a_n$ always diverges in that case.
Result
The series diverges because: