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Exercise 1504Calculus › Single-variable calculus › SequencesL2Added 2 Nov 2026

The alternating harmonic sums converge

#adjacent subsequences#alternating series#ln 2

Statement

Let Sn=∑k=1n(−1)k+1kS_n=\displaystyle\sum_{k=1}^n\frac{(-1)^{k+1}}k. Show that (S2n)(S_{2n}) and (S2n+1)(S_{2n+1}) are adjacent sequences, and deduce that (Sn)(S_n) converges.

Hint

Compute S2n+2−S2nS_{2n+2}-S_{2n}, S2n+3−S2n+1S_{2n+3}-S_{2n+1} and S2n+1−S2nS_{2n+1}-S_{2n}.

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Solution

Monotonicity.
S2n+2−S2n=12n+1−12n+2>0,S2n+3−S2n+1=−12n+2+12n+3<0.S_{2n+2}-S_{2n}=\frac1{2n+1}-\frac1{2n+2}>0,\qquad S_{2n+3}-S_{2n+1}=-\frac1{2n+2}+\frac1{2n+3}<0.
So (S2n)(S_{2n}) is increasing and (S2n+1)(S_{2n+1}) is decreasing.

Difference. S2n+1−S2n=12n+1→0S_{2n+1}-S_{2n}=\frac1{2n+1}\to0.

Adjacent sequences. They converge to the same limit ℓ\ell. Since both the even and the odd subsequences tend to ℓ\ell, the whole sequence (Sn)(S_n) converges to ℓ\ell.

Remark. The limit is ln⁡2≈0.6931\ln2\approx0.6931 (S2000≈0.6929S_{2000}\approx0.6929). The bounds S2n≤ln⁡2≤S2n+1S_{2n}\le\ln2\le S_{2n+1} give an error estimate of 12n+1\frac1{2n+1} for free.