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A convergent series ${\sum }_{n\ge 1}{u}_n$ has partial sums $S_N={\sum }_{n=1}^N{u}_n$ tending to a finite limit $S$. Explicitly, for every $\epsilon >0$ there is an integer $N_0$ such that $|S_N-S|<\epsilon$ whenever $N\ge N_0$. The convergence of the individual terms to zero is necessary but is not sufficient for convergence of the series.

For positive integers $n$, the summand is $a_n=\sin (\pi \sin (1/n))/n$. Since $0<\sin (1/n)<1$, these terms are positive. Apply the limit comparison test with the convergent p-series $\sum n^{-2}$:Each sine ratio tends to one. The limit comparison test says that two positive-term series whose term ratio tends to a finite positive number are either both convergent or both divergent. ThereforeThere is also a direct comparison test for series: $0\le \sin (\pi \sin (1/n))\le \pi \sin (1/n)\le \pi /n$, giving $0\le a_n\le \pi /n^2$.

A series adds the terms of a sequence in their specified order. Its infinite-sum notation is defined through finite partial sums:when the displayed limit of a sequence exists as a finite number. Otherwise the series is not a convergent series in the ordinary sense. Writing an infinite list of additions does not by itself assign it a finite sum. Cesàro summation is a different prescription, with its own definition.

Two positive series make the distinction concrete. For $a_n=1/n$, the individual terms tend to zero but the harmonic series has unbounded partial sums. For $a_n=1/[n(n+1)]$, partial fraction decomposition reveals a telescoping series whose partial sums tend to one. The useful question is how much all the remaining terms can contribute together, rather than only how small the next term looks.

For a convergent series with sum $S$ and $N$-th partial sum $S_N$, its series remainder is $R_N=S-S_N$. It equals the sum of the terms after position $N$. Its absolute value measures the error of replacing $S$ by $S_N$; it need not be positive when the series has signed sequence terms.

For a convergent series with sum $S$, the error from keeping only the first $N$ terms is the remaining sum $R_N={\sum }_{n>N}{a}_n$. This truncation error exists even with exact arithmetic. The zeta series truncation error gives explicit bounds for positive reciprocal-power terms.

For real $a$, the geometric-series expansion of $1/(e^x-1)$ gives an integrable series after multiplication by $\sin (ax)$: each absolute integral is bounded by $|a|{\int }_0^{\infty }x{e}^{-nx}dx=|a|/n^2$. Termwise integration is therefore justified and gives the convergent series ${\sum }_{n\ge 1}a/(a^2+n^2)$.

If a sequence in a metric space has ${\sum }_{j=1}^{\infty }d(x_{j+1},x_j)<\infty$, the triangle inequality gives $d(x_m,x_n)\le {\sum }_{j=n}^{m-1}d(x_{j+1},x_j)$ for $m>n$. The tail of this convergent series tends to zero and bounds every later pair, proving the Cauchy sequence condition. In particular, an increment bound $d(x_{j+1},x_j)\le C{r}^j$ with $C>0$ and $0<r<1$ gives $d(x_m,x_n)\le C{r}^n/(1-r)$ by the geometric series. Unlike checking a small latest step, this bounds all remaining movement.

This sufficient condition is not necessary. The real sequence $x_n=(-1{)}^n/n$ converges to zero and is therefore a Cauchy sequence, although $|x_{n+1}-x_n|=1/(n+1)+1/n$ has a divergent sum. Cancellation allows convergence despite infinite accumulated absolute values of the successive changes.

A sequence $(a_n)$ is summable when the series ${\sum }_n{a}_n$ is convergent.

A rule for assigning a value to some series through a specified limiting operation. It must say which series belong to its domain; different rules can have different domains. A regular summation method agrees with ordinary summation for every convergent series.

For a function with derivatives of every order at $a$, its Taylor series is the power seriesIts partial sums are Taylor polynomials. At $a=0$ it is called a Maclaurin series. Forming the coefficients, proving that its partial sums form a convergent sequence, and proving that their limit equals $f(x)$ are separate tasks. A smooth function provides all the coefficients, but a real analytic function is additionally represented locally by the resulting power series.

For the exponential function, every derivative of $e^x$ at zero equals one, givingHere equality holds for every real $x$: the Taylor remainder is bounded by $e^{|x|}|x{|}^{N+1}/(N+1)!$, which tends to zero for each fixed $x$. At $x=1$, the degree-five Taylor polynomial is $163/60\approx 2.71666667$, and the absolute difference from $e$ is less than $e/720$. The factorial denominator eventually overwhelms every fixed power, allowing local derivative information to reconstruct this function throughout the real line.

The geometric series shows why the region of validity matters. For $f(x)=1/(1-x)$, the Taylor series at zero is ${\sum }_{k\ge 0}{x}^k$, with radius of convergence one. For $x\ne 1$, its finite Taylor polynomial satisfies the exact identityAt $x=1/2$ and $N=5$, the remainder is $1/32$. At $x=0.9$, the same degree-five polynomial has value $4.68559$ and remainder $5.31441$, although the Taylor series still converges to $10$ there. Convergence of the infinite expression does not guarantee accuracy from a particular small number of terms. For $|x|>1$, even the terms fail to tend to zero, so this power series cannot represent the function. The standard expansions and the conditions for convergent series are discussed in OpenStax, working with Taylor series.

A stronger caution comes from the flat functionIt is a smooth function with $h^{(n)}(0)=0$ for every $n$, so every Taylor polynomial at zero is zero and its Taylor series converges everywhere to zero. Nevertheless $h(1/2)=e^{-4}>0$. Off zero, repeated differentiation produces a polynomial in $1/x$ times $e^{-1/x^2}$; the exponential decay makes each required limit at zero vanish. This explains how all the local coefficients can miss a positive function arbitrarily close to the centre. Real analyticity supplies the additional condition that rules out this behavior.

A telescoping series has partial sums in which internal contributions cancel. The basic pattern is $a_n=b_n-b_{n+1}$, for which finite summation givesOnly the two boundary terms survive. If $b_n$ tends to a finite limit of a sequence $L$, the series consequently sums to $b_1-L$. The cancellation is exact for each finite $N$; the infinite result then follows by a limit of a sequence.

For example, partial fraction decomposition givesThe first three terms are $1/2,1/6,1/12$, with partial sum $3/4$. The first five have partial sum $5/6$. Taking the limit of a sequence gives the concise answer

The appeal is that a long calculation becomes a boundary calculation. It works whenever the summand is a successive difference, not only for rational functions. For instance, $a_n=\log ((n+1)/n)=\log (n+1)-\log n$ gives partial sum $\log (N+1)$, so its series diverges despite the cancellation. A telescoping series therefore need not be a convergent series: one must still examine the surviving boundary term.

A telescoping sum cancels intermediate sequence terms, leaving boundary terms. For a finite sequence of real numbers $a_0,\dots ,a_n$, the displayed identity follows because every $a_1,\dots ,a_{n-1}$ occurs once with each sign. Applying it to an infinite series also requires checking the limit of the boundary terms; finite cancellation alone does not prove that the series is a convergent series.^[ref]

For $n\ge 1$, expanding the finite sum givesNo monotone sequence assumption is required. Consider the original sequence $4,9,2,7,5$. Its successive finite differences are $5,-7,5,-2$, soThe large intermediate rise and fall disappear from the net change. This is useful whenever one knows successive changes rather than every absolute value: adding the changes reconstructs the final value from the initial one. It also checks a cumulative-update calculation without requiring all increments to be positive.

A finite difference $\Delta a_{k-1}=a_k-a_{k-1}$ measures one discrete change. The identity ${\sum }_{k=1}^n\Delta a_{k-1}=a_n-a_0$ is the discrete counterpart of the fundamental theorem of calculus: accumulation of changes returns the endpoint difference. The related summation by parts transfers a finite difference from one factor of a product to another, with explicit boundary terms.

The term test for divergence says that a series cannot converge if its terms fail to tend to zero. Indeed, if its partial sums satisfy $S_n\to S$, then $a_n=S_n-S_{n-1}\to S-S=0$. This is a necessary condition for a convergent series.

The converse fails: $1/n\to 0$, but the harmonic series diverges. For a quick successful test, ${\sum }_{n=1}^{\infty }n/(n+1)$ diverges because its terms tend to one. The test gives no conclusion for $\sum 1/n$ or $\sum 1/[n(n+1)]$: both have terms tending to zero, yet only the second is a convergent series. The telescoping series calculation and the harmonic series block estimate resolve those cases.