Exercises — Telescoping series

What we do: just substitute. $a_n=\frac{1}{n}-\frac{1}{n+1}$. $a_1=1-\frac{1}{2},a_2=\frac{1}{2}-\frac{1}{3},a_3=\frac{1}{3}-\frac{1}{4},a_4=\frac{1}{4}-\frac{1}{5}.$ Why we don't add the hard way: the master formula already tells us the answer is first minus last. With $N=4$: $S_4=b_1-b_5=1-\frac{1}{5}=\boxed{\frac{4}{5}}.$ What it looks like: term $a_2$'s $+\frac{1}{2}$ kills term $a_1$'s $-\frac{1}{2}$; the $+\frac{1}{3},+\frac{1}{4}$ each get eaten too. Only $1$ (from the front) and $-\frac{1}{5}$ (from the back) survive.

(a) Not telescoping as written — it is a single blob, no difference visible. (It has no elementary telescoping split at all; leave it.) (b) Yes, but reversed sign order: $a_n=b_{n+1}-b_n$ with $b_n=\sqrt{n}$. That is still telescoping — the survivors just swap sign: $S_N=b_{N+1}-b_1=\sqrt{N+1}-1$. (c) Yes, textbook form: $b_n=\frac{1}{n}$, so $a_n=b_n-b_{n+1}$. ✓ Why (b) still counts: telescoping only needs consecutive cancellation; whether $b$ goes up or down is a sign bookkeeping detail, not a new phenomenon.

Step 1 — split. Want $\frac{1}{(2n-1)(2n+1)}=\frac{A}{2n-1}+\frac{B}{2n+1}$. Why partial fractions? It is the engine that turns one fraction into a difference we can telescope. Clear denominators: $1=A(2n+1)+B(2n-1)$. Set $n=\frac{1}{2}$: $1=A(2)\Rightarrow A=\frac{1}{2}$. Set $n=-\frac{1}{2}$: $1=B(-2)\Rightarrow B=-\frac{1}{2}$. $a_n=\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right).$ Step 2 — identify $b_n$. $b_n=\frac{1}{2n-1}$, and $b_{n+1}=\frac{1}{2n+1}$. Gap-1 telescope. ✓ Step 3 — partial sum. $S_N=\frac{1}{2}(b_1-b_{N+1})=\frac{1}{2}\left(1-\frac{1}{2N+1}\right).$ Step 4 — limit. $\frac{1}{2N+1}\to 0$, so $\sum =\frac{1}{2}(1-0)=\boxed{\frac{1}{2}}.$

Step 1 — split. $\frac{2}{n(n+2)}=\frac{A}{n}+\frac{B}{n+2}$. From $2=A(n+2)+Bn$: $n=0\Rightarrow A=1$; $n=-2\Rightarrow B=-1$. $a_n=\frac{1}{n}-\frac{1}{n+2}.$ Step 2 — spot the gap of 2. This is $b_n-b_{n+2}$ with $b_n=\frac{1}{n}$, not $b_n-b_{n+1}$. So two terms survive at each end. Step 3 — write out and collapse: $S_N=\left(1-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\cdots +\left(\frac{1}{N}-\frac{1}{N+2}\right).$ The $+\frac{1}{3}$ eats a later $-\frac{1}{3}$, etc. Survivors at the front: $1,\frac{1}{2}$. At the back: $-\frac{1}{N+1},-\frac{1}{N+2}$. $S_N=1+\frac{1}{2}-\frac{1}{N+1}-\frac{1}{N+2}.$ Step 4 — limit. Both tail terms $\to 0$: $\sum =1+\frac{1}{2}=\boxed{\frac{3}{2}}.$

Step 1 — identify $b_n$. Here $b_n=\frac{n}{n+1}$, and the term is literally $b_n-b_{n+1}$. No partial fractions needed. Step 2 — partial sum. $S_N=b_1-b_{N+1}=\frac{1}{2}-\frac{N+1}{N+2}.$ Step 3 — the crucial limit. Here $b_n=\frac{n}{n+1}\to 1$, not $0$ (see Limits of Sequences). So: $\sum =b_1-{\lim }_{n\to \infty }{b}_n=\frac{1}{2}-1=\boxed{-\frac{1}{2}}.$ Why the answer is negative: each pair steps down toward $1$, but the running total is anchored at $b_1=\frac{1}{2}$ minus the ceiling $1$. The leftover limit is the whole story here — the middle cancels, but the "last" survivor does not vanish.

Step 1 — split the log into a difference. Logs turn products/quotients into sums/differences: $\ln \frac{n(n+2)}{(n+1{)}^2}=\ln \frac{n}{n+1}-\ln \frac{n+1}{n+2}.$ Check by expanding: $\ln n-\ln (n+1)-\ln (n+1)+\ln (n+2)=\ln n+\ln (n+2)-2\ln (n+1)$ ✓. Step 2 — identify $b_n$. $b_n=\ln \frac{n}{n+1}$, term $=b_n-b_{n+1}$. Gap-1 telescope. Step 3 — partial sum. $S_N=b_1-b_{N+1}=\ln \frac{1}{2}-\ln \frac{N+1}{N+2}.$ Step 4 — limit. As $N\to \infty$, $\frac{N+1}{N+2}\to 1$, so $\ln \frac{N+1}{N+2}\to \ln 1=0$. Thus $\sum =\ln \frac{1}{2}-0=\boxed{-\ln 2}.$

Step 1 — same split. $\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$, so $b_n=\frac{1}{n}$. Step 2 — respect the starting index. The sum runs $n=3,4,5,\dots$. The first surviving term is $b_3$, not $b_1$: $S_N={\sum }_{n=3}^N\left(\frac{1}{n}-\frac{1}{n+1}\right)=b_3-b_{N+1}=\frac{1}{3}-\frac{1}{N+1}.$ Step 3 — limit. $\frac{1}{N+1}\to 0$, so ${\sum }_{n=3}^{\infty }=\frac{1}{3}-0=\boxed{\frac{1}{3}}.$ Sanity check: the full sum from $n=1$ is $1$; subtracting the first two terms $\frac{1}{2}+\frac{1}{6}=\frac{2}{3}$ gives $1-\frac{2}{3}=\frac{1}{3}$. ✓

Step 1 — the clever grouping. Notice $\frac{1}{n(n+1)(n+2)}=\frac{1}{2}\left[\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right].$ Verify: common denominator gives $\frac{1}{2}\cdot \frac{(n+2)-n}{n(n+1)(n+2)}=\frac{1}{2}\cdot \frac{2}{n(n+1)(n+2)}$ ✓. Why this shape? We want $b_n-b_{n+1}$. Setting $b_n=\frac{1}{n(n+1)}$ makes $b_{n+1}=\frac{1}{(n+1)(n+2)}$ — a perfect gap-1 telescope. Step 2 — partial sum. $S_N=\frac{1}{2}(b_1-b_{N+1})=\frac{1}{2}\left(\frac{1}{1\cdot 2}-\frac{1}{(N+1)(N+2)}\right).$ Step 3 — limit. The tail $\to 0$: $\sum =\frac{1}{2}\cdot \frac{1}{2}=\boxed{\frac{1}{4}}.$

Step 1 — engineer the difference. Using $n=(n+1)-1$: $\frac{n}{(n+1)!}=\frac{(n+1)-1}{(n+1)!}=\frac{n+1}{(n+1)!}-\frac{1}{(n+1)!}=\frac{1}{n!}-\frac{1}{(n+1)!}.$ Why this trick? We needed a telescoping shape; factorials shrink one step when the numerator matches $(n+1)$. So $b_n=\frac{1}{n!}$. Step 2 — partial sum. $S_N=b_1-b_{N+1}=\frac{1}{1!}-\frac{1}{(N+1)!}=1-\frac{1}{(N+1)!}.$ Step 3 — limit. $\frac{1}{(N+1)!}\to 0$ (factorials blow up), so $\sum =\boxed{1}.$

Part A — the gap-2 theorem. Write the partial sum and match survivors: $S_N={\sum }_{n=1}^N(b_n-b_{n+2})=(b_1+b_2+\cdots +b_N)-(b_3+b_4+\cdots +b_{N+2}).$ Everything from $b_3$ to $b_N$ appears in both brackets and cancels. Front survivors: $b_1,b_2$. Back survivors: $-b_{N+1},-b_{N+2}$. $S_N=b_1+b_2-b_{N+1}-b_{N+2}.$ Take $N\to \infty$: since $b_n\to L$, both tail terms $\to L$, giving ${\sum }_{n=1}^{\infty }(b_n-b_{n+2})=b_1+b_2-2L.■$ Part B — apply to $b_n=\frac{1}{n}$ (so $L=0$). Then $b_n-b_{n+2}=\frac{1}{n}-\frac{1}{n+2}=\frac{2}{n(n+2)}$, and the theorem gives ${\sum }_{n=1}^{\infty }\frac{2}{n(n+2)}=b_1+b_2-0=1+\frac{1}{2}=\frac{3}{2},$ hence $\sum _{n=1}^{\infty }\frac{1}{n(n+2)}=\frac{3}{4}$. ✓ (Matches the parent note — the general theorem contains that example.)