Series Calculator

Analyze convergence, compute sums, and expand Taylor/Maclaurin series with step-by-step solutions

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∑Math Input
sum of 1/n^2 from n=1 to infinity
does sum of (-1)^n / n converge?
Taylor series of sin(x) at x = 0
sum of (2/3)^n from n=0 to infinity

What is a Series?

A series is the sum of the terms of a sequence. An infinite series takes the form:

∑n=1∞an=a1+a2+a3+⋯\sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + \cdots

The partial sums are SN=∑n=1NanS_N = \sum_{n=1}^{N} a_n. If the sequence of partial sums converges to a finite limit SS, we say the series converges and ∑n=1∞an=S\sum_{n=1}^{\infty} a_n = S. Otherwise, the series diverges.

Geometric Series: The series ∑n=0∞arn\sum_{n=0}^{\infty} ar^n converges to a1−r\frac{a}{1-r} when ∣r∣<1|r| < 1.

p-Series: The series ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p} converges when p>1p > 1 and diverges when p≤1p \leq 1.

Power Series: A series of the form ∑n=0∞cn(x−a)n\sum_{n=0}^{\infty} c_n (x - a)^n that represents a function within its radius of convergence.

Taylor Series: The power series expansion of f(x)f(x) about x=ax = a:

f(x)=∑n=0∞f(n)(a)n!(x−a)nf(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n

When a=0a = 0, this is called a Maclaurin series.

How to Determine Convergence

Divergence Test (nth-term test)

If lim⁡n→∞an≠0\lim_{n \to \infty} a_n \neq 0, the series diverges. Note: if the limit is 0, the test is inconclusive.

Ratio Test

Compute L=lim⁡n→∞∣an+1an∣L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|:

  • If L<1L < 1: converges absolutely
  • If L>1L > 1: diverges
  • If L=1L = 1: inconclusive

Root Test

Compute L=lim⁡n→∞∣an∣nL = \lim_{n \to \infty} \sqrt[n]{|a_n|}. Same conclusion rules as the Ratio Test.

Integral Test

If f(n)=anf(n) = a_n where ff is positive, continuous, and decreasing for x≥1x \geq 1:
∑n=1∞an converges  ⟺  ∫1∞f(x) dx converges\sum_{n=1}^{\infty} a_n \text{ converges} \iff \int_1^{\infty} f(x)\,dx \text{ converges}

Comparison Test

If 0≤an≤bn0 \leq a_n \leq b_n for all nn:

  • If ∑bn\sum b_n converges, then ∑an\sum a_n converges
  • If ∑an\sum a_n diverges, then ∑bn\sum b_n diverges

Alternating Series Test (Leibniz Test)

The alternating series ∑(−1)nbn\sum (-1)^n b_n converges if:

  1. bn>0b_n > 0 for all nn
  2. bnb_n is decreasing
  3. lim⁡n→∞bn=0\lim_{n \to \infty} b_n = 0

Common Taylor/Maclaurin Series

FunctionMaclaurin SeriesRadius
exe^x∑n=0∞xnn!\sum_{n=0}^{\infty} \frac{x^n}{n!}∞\infty
sin⁡x\sin x∑n=0∞(−1)nx2n+1(2n+1)!\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}∞\infty
cos⁡x\cos x∑n=0∞(−1)nx2n(2n)!\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}∞\infty
11−x\frac{1}{1-x}∑n=0∞xn\sum_{n=0}^{\infty} x^n11
ln⁡(1+x)\ln(1+x)∑n=1∞(−1)n+1xnn\sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n}11

Choosing the Right Test

TestBest ForKey Indicator
DivergenceQuick eliminationTerms clearly don't approach 0
RatioFactorials, exponentialsn!n! or rnr^n in terms
Rootnth powersan=[f(n)]na_n = [f(n)]^n
IntegralSimple decreasing functionsan=f(n)a_n = f(n) easily integrated
ComparisonTerms resemble known seriesLooks like p-series or geometric
AlternatingSign-alternating series(−1)n(-1)^n factor

Common Mistakes to Avoid

  • Misusing the Divergence Test: If lim⁡an=0\lim a_n = 0, this does NOT prove convergence. The harmonic series ∑1/n\sum 1/n diverges even though 1/n→01/n \to 0.
  • Applying Ratio Test when L = 1: When the ratio limit equals 1, the test gives no information. You must use a different test.
  • Confusing absolute and conditional convergence: A series can converge conditionally (like the alternating harmonic series) without converging absolutely.
  • Wrong radius of convergence: Don't forget to check the endpoints separately when finding the interval of convergence.
  • Taylor series remainder: The Taylor polynomial is only an approximation; for finite terms, there is a remainder term whose bound matters for accuracy.

Examples

Step 1: Apply the Ratio Test: an+1an=(n+1)/2n+1n/2n=n+12n\frac{a_{n+1}}{a_n} = \frac{(n+1)/2^{n+1}}{n/2^n} = \frac{n+1}{2n}
Step 2: L=lim⁡n→∞n+12n=12<1L = \lim_{n \to \infty} \frac{n+1}{2n} = \frac{1}{2} < 1, so the series converges
Step 3: To find the sum, use the formula ∑n=1∞nxn=x(1−x)2\sum_{n=1}^{\infty} nx^n = \frac{x}{(1-x)^2} with x=12x = \frac{1}{2}: 1/2(1/2)2=2\frac{1/2}{(1/2)^2} = 2
Answer: 22

Step 1: Start with the geometric series: 11−t=∑n=0∞tn\frac{1}{1-t} = \sum_{n=0}^{\infty} t^n for ∣t∣<1|t| < 1
Step 2: Substitute t=−x2t = -x^2: 11+x2=11−(−x2)=∑n=0∞(−x2)n\frac{1}{1+x^2} = \frac{1}{1-(-x^2)} = \sum_{n=0}^{\infty} (-x^2)^n
Step 3: Simplify: ∑n=0∞(−1)nx2n=1−x2+x4−x6+⋯\sum_{n=0}^{\infty} (-1)^n x^{2n} = 1 - x^2 + x^4 - x^6 + \cdots for ∣x∣<1|x| < 1
Answer: ∑n=0∞(−1)nx2n\sum_{n=0}^{\infty} (-1)^n x^{2n}, valid for ∣x∣<1|x| < 1

Step 1: This is an alternating series with bn=1nb_n = \frac{1}{\sqrt{n}}
Step 2: Check: bn>0b_n > 0 ✓, bnb_n is decreasing ✓, lim⁡n→∞bn=0\lim_{n \to \infty} b_n = 0 ✓
Step 3: By the Alternating Series Test, the series converges (conditionally, since ∑1n\sum \frac{1}{\sqrt{n}} diverges as a p-series with p=1/2<1p = 1/2 < 1)
Answer: The series converges conditionally

Frequently Asked Questions

A series converges if its partial sums approach a finite number as you add more terms. A series diverges if the partial sums grow without bound or oscillate without settling on a value.

Taylor series are used to approximate complicated functions with polynomials, making them easier to compute, differentiate, or integrate. They are fundamental in physics, engineering, and numerical analysis for approximating functions near a specific point.

The radius of convergence R is the distance from the center of a power series within which the series converges. For |x - a| < R the series converges absolutely, for |x - a| > R it diverges, and at |x - a| = R you must check endpoints individually.

No. The harmonic series, which is the sum of 1/n from n=1 to infinity, diverges. Even though the terms approach zero, they do not decrease fast enough for the sum to remain finite. This is a classic example showing that terms going to zero is necessary but not sufficient for convergence.

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