1.3 Summation of series

The arithmetic series 1+2+…+n equals n(n+1)/2. The geometric series a+ar+…+ar^{n−1} equals a(1−r^n)/(1−r) when r≠1, and tends to a/(1−r) when |r|<1.

Index the first and last terms carefully. For an infinite series, convergence requires the terms to tend to zero; a ratio with |r|≥1 does not give a finite sum.

The sum 3+6+12+24 is a=3, r=2, n=4, so it is 3(1−2^4)/(1−2)=45. The corresponding infinite series diverges because |r|>1.

Do not use the infinite formula for a finite or divergent series, and do not confuse the number of terms with the final index.

A telescoping or method-of-differences series is written as differences such as u_r−u_{r+1}, so most intermediate terms cancel when the terms are added.

Expand the first few and last few terms before simplifying. The answer is determined by the uncancelled boundary terms; if the cancellation pattern changes at a singular index, split the sum.

Σ_{r=1}^{n}(1/r−1/(r+1)) = 1−1/(n+1), because every middle reciprocal cancels.

Do not cancel across a plus sign or ignore the final boundary term; write at least three terms before jumping to the pattern.

For an infinite series to converge, its terms must tend to zero and the partial sums must approach a finite limit. A geometric series converges when |r|<1; otherwise its partial sums do not settle to a finite sum.

The term test is necessary but not sufficient for every series: terms tending to zero does not by itself prove convergence. Use an appropriate comparison, ratio, integral or known-series test when required.

Σ(1/2)^n converges because the ratio has magnitude below one. Σ1/n has terms tending to zero but still diverges, so checking only the terms is insufficient.

“The terms get small” is not a complete convergence argument, and a finite partial sum is not the value of an infinite series.