Infinite Series and Convergence

An infinite series is a limit of partial sums

For a sequence $a_1,a_2,\ldots$, form the finite running total

If $S_N$ tends to a finite number $S$, write $\sum_{n=1}^\infty a_n=S$. If there is no finite partial-sum limit, the series diverges. The index $n$ selects a term; the upper index $N$ records how many terms have been included in this definition.

For the terms $1/2,1/4,1/8,\ldots$, the first partial sums are $1/2,3/4,7/8,\ldots$. The terms tend to zero while the sums tend to one. An infinite series does not mean performing infinitely many additions as a single finite operation. It means taking this limit.

Adding, removing or changing a finite prefix does not change convergence, but it generally changes the sum. That is why starting indices matter when a numerical answer is required.

Geometric series

A geometric series has terms $a,ar,ar^2,\ldots$, where $a$ is the first term and $r$ is the fixed ratio. With exactly $N$ terms, starting at exponent zero,

Multiply this finite sum by $r$ and subtract it from the original. All the interior terms cancel:

For $r\ne1$,

If $\lvert r\rvert<1$, then $r^N\to0$, which proves

For a non-zero first term, $\lvert r\rvert\ge1$ prevents convergence because the terms do not tend to zero. The zero series $a=0$ is an exception to that wording: all its terms are zero. When $r=1$, the finite sum is $Na$, not a quotient with denominator zero.

Calculate a geometric sum and its remaining tail

For $1+1/3+1/9+\cdots$, use $a=1$ and $r=1/3$. The sum is $3/2$, while the first three terms total $13/9$. The omitted tail is

For any convergent geometric series, the absolute error after $N$ terms is $\lvert a\rvert\lvert r\rvert^N/\lvert1-r\rvert$. This remains valid for negative $r$, even though the partial sums then alternate around their limit.

Read the area illustration precisely

The diagram uses a different ratio. Relative to the large triangle, the successive black triangles have areas $1/4,1/16,1/64,\ldots$. Their side lengths halve, so their areas are multiplied by $1/4$ at each step.

A quarter-area sequence inside one triangle

Thus the black regions occupy $\sum_{n=1}^\infty(1/4)^n=1/3$ of the full triangle. They do not fill its whole area. The first black region has area $1/4$, not $1$, and the index begins at one. Keeping those details explicit prevents a picture from suggesting a different sum.

Telescoping series

Some terms can be rewritten as differences whose neighbouring contributions cancel. For $n\ge1$,

Write a finite partial sum before taking the limit:

The surviving boundary terms give

Consequently the infinite sum is $1$, and the exact omitted tail is $1/(N+1)$. For instance, after $99$ terms the tail is $0.01$.

Do not justify this by splitting the infinite series into $\sum1/n-\sum1/(n+1)$. Each of those series diverges. The valid cancellation takes place in finite partial sums, where both sums contain the same prescribed number of terms.

Terms must approach zero, but that is not enough

If $S_N\to S$, then $a_N=S_N-S_{N-1}\to S-S=0$. This proves the necessary condition: a convergent series must have terms tending to zero.

The converse fails. For the harmonic series $\sum_{n=1}^\infty1/n$, the individual terms tend to zero, but the running total does not stay bounded. Group the terms after the first into blocks:

The block from $n=2^{k-1}+1$ through $2^k$ has $2^{k-1}$ terms, each at least $1/2^k$. Each such block contributes at least $1/2$. Adding more blocks therefore makes the partial sums arbitrarily large.

This explains the limitation of the nth-term test. A non-zero term limit, or no term limit, proves divergence. A zero term limit leaves the question open. The infinite-series reference develops these partial-sum arguments.

p-series benchmarks

A p-series is $\sum_{n=1}^\infty1/n^p$ for a fixed real exponent $p$. It converges when $p>1$ and diverges when $p\le1$. The integral test supplies a useful proof for positive decreasing powers; for $p\le0$, the terms already fail to approach zero.

Thus $\sum1/n^2$ converges, $\sum1/n$ diverges, and $\sum1/\sqrt n$ diverges even though its terms tend to zero. A larger positive exponent makes these terms smaller, so the direction of the threshold is consistent with comparison.

The exact value $\sum_{n=1}^\infty1/n^2=\pi^2/6$ is a separate result. The convergence test establishes that a finite sum exists, not that numerical identity. Most convergence problems require the classification and its reason, not a closed-form sum.

Absolute and conditional convergence

A series converges absolutely when $\sum\lvert a_n\rvert$ converges. Absolute convergence implies convergence of the original signed series. A series converges conditionally when the signed series converges but the absolute-value series does not.

The alternating harmonic series

is conditionally convergent. Its magnitudes decrease to zero, so the alternating series test gives convergence. Replacing all signs by positive signs produces the divergent harmonic series. The value $\ln2$ can be obtained from the logarithmic power series, but that identification is separate from the alternating-series convergence test.

An alternating sign pattern is not enough on its own: $1-1+1-1+\cdots$ has partial sums alternating between $1$ and $0$. Its terms do not decay. For conditionally convergent series, keep the given ordering; unrestricted rearrangement can alter the limit. Absolute convergence is what permits arbitrary rearrangement without changing the sum.

Related topics

Use Convergence Tests for Series when a series is not immediately geometric, telescoping or a standard p-series. That page also turns convergence information into useful bounds on the omitted tail.

Practice questions

Sources and further reading

• OpenStax Calculus Volume 2: Infinite Series. Partial sums, geometric and telescoping series, harmonic divergence.
• OpenStax: Alternating Series. Absolute and conditional convergence, ordering and error estimates.
• OpenStax: Divergence and Integral Tests. p-series convergence via integral comparison.
• Wolfram Language: Calculus. Independent symbolic calculations and limit verification.