Exercises — Series — partial sums, convergence definition

WHAT is $a$ and $r$? The template is $\sum a{r}^{n-1}$. Matching term by term: the constant multiplying the powers is $a=5$; the base being raised to $n-1$ is $r=\frac{2}{3}$.

WHY does the sign/size of $r$ matter? The geometric series converges exactly when $|r|<1$, because the partial sum is $s_N=\frac{a(1-r^N)}{1-r}$ and only $r^N$ depends on $N$. When $|r|<1$, $r^N\to 0$; the running total stops moving.

Here $|r|=\frac{2}{3}<1$, so it converges to $S=\frac{a}{1-r}=\frac{5}{1-\frac{2}{3}}=\frac{5}{\frac{1}{3}}=15.$

WHAT makes it telescoping? Each term is a difference of two pieces of the same shape $\frac{1}{n}$ evaluated at shifted indices ($n$ and $n+2$). When you add consecutive terms, the subtracted piece of one term reappears as the positive piece two terms later, so it cancels.

WHICH pieces survive? Because the gap is $2$ (not $1$), the cancellation leaves two leading positives and two trailing negatives: $s_N=\underset{\text{first two }+}{\underbrace{1+\frac{1}{2}}}-\underset{\text{last two }-}{\underbrace{\frac{1}{N+1}-\frac{1}{N+2}}}.$ The middle all cancels. (We compute the actual value in L3·Q2.)

Rewrite into $a{r}^{n-1}$ form. $\frac{3}{4^n}=3\cdot 4^{-n}=3\cdot 4^{-1}\cdot 4^{-(n-1)}=\frac{3}{4}{\left(\frac{1}{4}\right)}^{n-1}$. So $a=\frac{3}{4}$, $r=\frac{1}{4}$. (Check by $n=1$: $\frac{3}{4^1}=\frac{3}{4}$ ✓.)

Test convergence. $|r|=\frac{1}{4}<1$, so it converges. $S=\frac{a}{1-r}=\frac{\frac{3}{4}}{1-\frac{1}{4}}=\frac{\frac{3}{4}}{\frac{3}{4}}=1.$

WHY reach for the $n$-th Term Test? It answers one cheap question first: do the terms even shrink to $0$? If they don't, the series can't converge (a necessary condition), and we're done.

Compute the limit of the term. Divide top and bottom by $n^2$: ${\lim }_{n\to \infty }\frac{2n^2+1}{5n^2+n}={\lim }_{n\to \infty }\frac{2+\frac{1}{n^2}}{5+\frac{1}{n}}=\frac{2}{5}\ne 0.$ Since $a_n\to \frac{2}{5}\ne 0$, by the n-th Term Test for Divergence the series diverges.

Set it up as a sum. $0.\overline{7}=\frac{7}{10}+\frac{7}{100}+\frac{7}{1000}+\cdots ={\sum }_{n=1}^{\infty }\frac{7}{10}{\left(\frac{1}{10}\right)}^{n-1}$, so $a=\frac{7}{10}$, $r=\frac{1}{10}$.

Sum it. $|r|=\frac{1}{10}<1$: $S=\frac{\frac{7}{10}}{1-\frac{1}{10}}=\frac{\frac{7}{10}}{\frac{9}{10}}=\frac{7}{9}.$ So $0.\overline{7}=\frac{7}{9}$ — repeating decimals are just convergent geometric series in disguise.

Identify the ratio. This is geometric with $a=1$ and $r=x-2$.

Convergence condition. We need $|r|<1$, i.e. $|x-2|<1$, i.e. $-1<x-2<1$, i.e. $1<x<3.$ Look at Figure 1: the shaded interval $(1,3)$ is exactly the strip where the ratio stays inside $(-1,1)$. Outside it the powers $r^{n-1}$ don't shrink and $s_N$ has no limit.

Endpoint check (never skip these!):

• $x=3\Rightarrow r=1$: $s_N=N\to \infty$, diverges.
• $x=1\Rightarrow r=-1$: $s_N$ bounces $1,0,1,0,\dots$, no limit, diverges.

Sum inside the interval. $S=\frac{a}{1-r}=\frac{1}{1-(x-2)}=\frac{1}{3-x}$ for $1<x<3$. (This is a first taste of Power series and radius of convergence.)

Write out $s_N$ carefully. Line up the two shifted lists:

Compute the partial sums. Terms are $1,-1,1,-1,\dots$ $s_1=1,s_2=0,s_3=1,s_4=0,\dots$ So $s_N=1$ for odd $N$ and $s_N=0$ for even $N$.

Look at Figure 2: the running total forever hops between $1$ and $0$. It never approaches a single value, so ${\lim }_{N\to \infty }{s}_N$ does not exist.

Conclusion. Divergence does not require blowing up to infinity — bouncing between two heights without settling is also divergence. (Consistent with the $n$-th Term Test too: $a_n=(-1{)}^{n-1}\nrightarrow 0$.)

WHY may we split it? If two series $\sum b_n$ and $\sum c_n$ both converge, then $\sum (b_n+c_n)$ converges to the sum of the two limits — because the partial sums add: $s_N=\sum b_n+\sum c_n$, and the limit of a sum is the sum of the limits.

Piece 1 — geometric. ${\sum }_{n=1}^{\infty }\frac{1}{2^n}=\sum \frac{1}{2}(\frac{1}{2}{)}^{n-1}$, $a=r=\frac{1}{2}$, $|r|<1$: converges to $\frac{\frac{1}{2}}{1-\frac{1}{2}}=1$.

Piece 2 — telescoping. ${\sum }_{n=1}^{\infty }\frac{1}{n(n+1)}$: from the parent note this telescopes to $s_N=1-\frac{1}{N+1}\to 1$.

Combine. Both converge, so the total converges to $S=1+1=2.$

Analyse each piece. $\sum \frac{1}{3^n}$ is geometric with $r=\frac{1}{3}$: converges (to $\frac{1}{2}$). But $\sum \frac{1}{n}$ is the Harmonic series: diverges.

WHY the whole thing diverges. Write $s_N=A_N+B_N$ where $A_N\to \frac{1}{2}$ (bounded) and $B_N={\sum }_{n=1}^N\frac{1}{n}\to \infty$. A convergent-plus-divergent sum is divergent: if $s_N$ converged, then $B_N=s_N-A_N$ would be a difference of two convergent sequences, hence convergent — contradicting that $\sum \frac{1}{n}$ diverges. So the series diverges.

(Moral: one bad apple spoils the sum. You can only add limits when both exist.)

Step 1 — partial fractions (WHY: to force a difference that cancels). Solve $\frac{2}{(2n-1)(2n+1)}=\frac{A}{2n-1}+\frac{B}{2n+1}$. Multiply out: $2=A(2n+1)+B(2n-1)$. Setting $n=\frac{1}{2}$ gives $A=1$; setting $n=-\frac{1}{2}$ gives $B=-1$. So $\frac{2}{(2n-1)(2n+1)}=\frac{1}{2n-1}-\frac{1}{2n+1}.$

Step 2 — partial sum telescopes. $s_N=\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\cdots +\left(\frac{1}{2N-1}-\frac{1}{2N+1}\right)=1-\frac{1}{2N+1}.$

Step 3 — limit. $s_N=1-\frac{1}{2N+1}\to 1$. The series converges to $1$.

(a) WHY the sum equals $\lim s_N$. By definition the series' value is the limit of its partial sums. Divide top and bottom by $N$: ${\lim }_{N\to \infty }\frac{3N}{N+2}={\lim }_{N\to \infty }\frac{3}{1+\frac{2}{N}}=3.$ So it converges to $3$.

(b) Recover the terms. From the parent note, $a_N=s_N-s_{N-1}$ (the running total's jump). For $N\ge 2$:

Identify $r$. With $r=\frac{p}{p+1}$ and first term (at $n=1$) $a=\frac{p}{p+1}$, it's geometric.

When does $|r|<1$? Since $p>0$, we have $0<\frac{p}{p+1}<1$ always (numerator smaller than denominator, both positive). So it converges for every $p>0$.

Sum. $S=\frac{a}{1-r}=\frac{\frac{p}{p+1}}{1-\frac{p}{p+1}}=\frac{\frac{p}{p+1}}{\frac{1}{p+1}}=p.$ A clean result: the whole series sums to exactly $p$.