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Alternating Series Calculator

Sum an alternating series, check the Leibniz conditions and report the error bound.

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Σ(-1)^(n-1) a_n ， |R_N| ≤ a_{N+1}
without sign, e.g. 1/n
number of terms

📖 Tutorial | Alternating Series

1. Theorem
Leibniz test: if a_n is positive, decreasing and tends to 0, the alternating series Σ(-1)^(n-1) a_n converges and |S−S_N| ≤ a_{N+1}.
2. Symbols
a_npositive part
S_Npartial sum
Ssum
|S−S_N|≤a_{N+1}error bound
3. How it works
  • Sum terms with alternating signs;
  • Check a_{N+1}≤a_N (decreasing);
  • Check a_{N+1}≈0;
  • Report the bound a_{N+1}.
4. Steps
  1. Enter the positive part a_n, e.g. 1/n;
  2. Enter N, e.g. 100;
  3. Click Calculate.
5. Example
Example: alternating harmonic Σ(-1)^(n-1)/n.
Solution: a_n=1/n decreases to 0. S_100≈0.698, error bound a_101≈0.0099; the sum ln2≈0.6931 matches.
6. Pitfalls
Require decreasing to 0 before using the error bound;
Alternating series may be conditionally convergent;
The bound holds only under the Leibniz conditions.

❓ FAQ | Alternating Series

Do all alternating series converge?
No—only when a_n decreases to 0.
How to use the error bound?
Choose N so a_{N+1}<ε for accuracy ε.
What is conditional convergence?
Converges but |a_n| diverges, e.g. alternating harmonic.
Why does S_N oscillate?
Alternating partial sums bracket the true value.
What if a_n is not decreasing?
The Leibniz bound does not apply; use another method.