Q.Write the first 4 terms of the logarithmic series

A finite sum of real numbers is always well-defined, but an infinite sum $a_1+a_2+a_3+\cdots$ needs a precise meaning, since one cannot literally "add infinitely many numbers" the way one adds finitely many.

Convergence of a sequence. A sequence $(a_n)$ converges to a limit $a$ — written $\lim_{n\to\infty}a_n=a$ — if $a_n$ gets and stays arbitrarily close to $a$ as $n$ grows without bound. Not every sequence converges (e.g. $1,0,1,0,\ldots$ does not settle near any single value), but if a sequence does converge, its limit is unique.

Convergence of a series. For a series $\sum_{n=1}^\infty a_n$, form the partial-sum sequence $s_n=a_1+a_2+\cdots+a_n$. The series is convergent with sum $s$ exactly when the partial sums converge, $\lim_{n\to\infty}s_n=s$; we then write $\sum_{n=1}^\infty a_n=s$. (The motivating cake experiment — repeatedly halving a cake between two plates — makes $\sum \frac1{2^n}=1$ intuitive this way: the partial sums $\frac12,\frac34,\frac78,\ldots$ visibly close in on $1$.) Ordinary finite-series algebra (freely regrouping terms) can give contradictory "answers" if misapplied to a non-convergent series such as $1-1+1-1+\cdots$ — a caution the chapter states explicitly.

Standard convergent series (each valid on a stated range of $x$):

• Geometric: $\sum_{n=0}^\infty x^n=\dfrac1{1-x}$ for $|x|<1$; with variants $\sum(-1)^nx^n=\dfrac1{1+x}$ and $\sum(2x)^n=\dfrac1{1-2x}$ (valid $|x|<\tfrac12$).
• Arithmetico-geometric: $\sum_{n=1}^\infty (a+(n-1)d)r^{n-1} = \dfrac a{1-r}+\dfrac{dr}{(1-r)^2}$ for $-1<r<1$.
• Telescoping infinite series: the same cancel-the-middle idea as the finite case, letting $n\to\infty$ in the partial sum (e.g. $\sum \frac1{n^2+5n+6}=\frac13$, from $\frac1{n+2}-\frac1{n+3}$). …