Exercises — Sigma (Σ) notation — evaluating, telescoping sums

WHAT we do: substitute each integer $k=2,3,4,5$ into the general term $2k-1$. $k=2:3,k=3:5,k=4:7,k=5:9.$ WHY: $\sum$ is only shorthand for "add the general term for every integer from the lower to the upper limit." Add: $3+5+7+9=24$. Sanity check on the count: $n-m+1=5-2+1=4$ terms — and yes we wrote four numbers. ✓

WHAT: count integers from $-3$ to $7$ inclusive. WHY the "+1": subtracting endpoints ($7-(-3)=10$) loses one term, so add it back. $n-m+1=7-(-3)+1=10+1=11.$ Quick check: list them $-3,-2,-1,0,1,2,3,4,5,6,7$ — eleven values. ✓

WHAT: the general term is the constant $7$; it does not depend on $k$. WHY: a constant added $n$ times is $nc$. Here $n=12$ terms, each equal to $7$. ${\sum }_{k=1}^{12}7=12\cdot 7=84.$

Step 1 (split by linearity). $\sum _{k=1}^{30}(4k+5)=4\sum _{k=1}^{30}k+\sum _{k=1}^{30}5.$ Why: $\sum$ distributes over $+$ and pulls constants out — one hard sum becomes two known pieces. Step 2 (standard formula + constant). ${\sum }_{k=1}^{30}k=\frac{30\cdot 31}{2}=465,{\sum }_{k=1}^{30}5=30\cdot 5=150.$ Step 3. $4(465)+150=1860+150=2010.$

Step 1 (split). $3\sum k^2-\sum k$. Step 2 (plug in formulas). $3\cdot \frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}=\frac{n(n+1)(2n+1)}{2}-\frac{n(n+1)}{2}.$ Step 3 (common denominator, factor $\frac{n(n+1)}{2}$). $\frac{n(n+1)}{2}[(2n+1)-1]=\frac{n(n+1)}{2}\cdot 2n=n^2(n+1).$ So $\sum _{k=1}^n(3k^2-k)={\mathbf{n}}^2(\mathbf{n}+1).$ Check $n=3$: direct sum $=(3\cdot 1-1)+(3\cdot 4-2)+(3\cdot 9-3)=2+10+24=36$; formula $=9\cdot 4=36$. ✓

WHAT: we only have a formula that starts at $k=1$. So compute the whole run to $20$ and subtract the missing head $k=1\dots 5$. WHY: ${\sum }_{k=6}^{20}={\sum }_{k=1}^{20}-{\sum }_{k=1}^5$ — the "start late" trick. ${\sum }_{k=1}^{20}k=\frac{20\cdot 21}{2}=210,{\sum }_{k=1}^{5}k=\frac{5\cdot 6}{2}=15.$ ${\sum }_{k=6}^{20}k=210-15=195.$

Step 1 (partial fractions). Write $\frac{1}{k(k+1)}=\frac{A}{k}+\frac{B}{k+1}$; clearing denominators $1=A(k+1)+Bk$. Set $k=0\Rightarrow A=1$; set $k=-1\Rightarrow B=-1$. So $\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}=f(k)-f(k+1),f(k)=\frac{1}{k}.$ Step 2 (telescope). By the anchor formula, sum $=f(1)-f(n+1)=1-\frac{1}{n+1}=\frac{n}{n+1}.$ Step 3 (evaluate). At $n=50$: $\frac{50}{51}\approx 0.98039$. So $\sum _{k=1}^{50}\frac{1}{k(k+1)}=\frac{50}{51}.$ Look at figure s01: the blue bars are the terms, and the red/green arrows show each $-\frac{1}{k+1}$ cancelling the next $+\frac{1}{k+1}$; only the far-left $1$ and far-right $-\frac{1}{n+1}$ survive.

Step 1 (partial fractions). $\frac{1}{k(k+2)}=\frac{1}{2}(\frac{1}{k}-\frac{1}{k+2}).$ Why the $\frac{1}{2}$: the gap between factors is $2$, and the split leaves a $\frac{1}{2}$ scale. Step 2 (the cancellation is delayed). Because the tail is $\frac{1}{k+2}$ (not $\frac{1}{k+1}$), each negative piece cancels the positive piece two rows later, not the next one. Surviving heads: $\frac{1}{1}+\frac{1}{2}$. Surviving tails: $-\frac{1}{n+1}-\frac{1}{n+2}$. Two survivors each end. Step 3. ${\sum }_{k=1}^n\frac{1}{k(k+2)}=\frac{1}{2}(1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2})=\frac{1}{2}(\frac{3}{2}-\frac{1}{n+1}-\frac{1}{n+2}).$ Evaluate $n=4$: $\frac{1}{2}(\frac{3}{2}-\frac{1}{5}-\frac{1}{6})=\frac{1}{2}\left(\frac{45-6-5}{30}\right)=\frac{1}{2}\cdot \frac{34}{30}=\frac{17}{30}.$ So $\sum _{k=1}^4\frac{1}{k(k+2)}=\frac{17}{30}.$ Figure s02 shows the "gap-2" offset: each blue term reaches across an empty slot before it meets its cancelling partner.

Way 1 (telescope). The term is $f(k+1)-f(k)$ with $f(k)=k^2$ — a reversed difference, so it collapses to $f(n+1)-f(1)=(n+1{)}^2-1$. Way 2 (expand then use formulas). $(k+1{)}^2-k^2=2k+1$, so ${\sum }_{k=1}^n(2k+1)=2\cdot \frac{n(n+1)}{2}+n=n(n+1)+n=n^2+2n=(n+1{)}^2-1.$ Both give $(\mathbf{n}+1{)}^2-1=n^2+2n$. ✓ (At $n=5$: $35$ both ways.)

Step 1 (design the partial fraction). The factors differ by $2$, so expect a $\frac{1}{2}$ scale: $\frac{1}{(2k-1)(2k+1)}=\frac{1}{2}(\frac{1}{2k-1}-\frac{1}{2k+1}).$ Check: $\frac{1}{2}\cdot \frac{(2k+1)-(2k-1)}{(2k-1)(2k+1)}=\frac{1}{2}\cdot \frac{2}{(2k-1)(2k+1)}$ ✓. Step 2 (telescope with $f(k)=\frac{1}{2k-1}$). Now $a_k=\frac{1}{2}[f(k)-f(k+1)]$ since $2(k+1)-1=2k+1$. Sum $=\frac{1}{2}[f(1)-f(n+1)]=\frac{1}{2}(\frac{1}{1}-\frac{1}{2n+1}).$ Step 3 (simplify). ${\sum }_{k=1}^n\frac{1}{(2k-1)(2k+1)}=\frac{1}{2}\cdot \frac{2n}{2n+1}=\frac{n}{2n+1}.$ Evaluate $n=10$: $\frac{10}{21}$. So the answer is $\frac{\mathbf{n}}{2n+1}$, giving $\frac{10}{21}$ at $n=10$.

Step 1 (spot the "shrink the gap" pattern). The clean trick: write it as a difference of the neighbouring product $\frac{1}{k(k+1)}$: $\frac{1}{k(k+1)}-\frac{1}{(k+1)(k+2)}=\frac{(k+2)-k}{k(k+1)(k+2)}=\frac{2}{k(k+1)(k+2)}.$ WHY this move: the numerator became a constant $2$, so dividing by $2$ isolates our term as a clean difference. Hence with $g(k)=\frac{1}{k(k+1)}$, $\frac{1}{k(k+1)(k+2)}=\frac{1}{2}[g(k)-g(k+1)].$ Step 2 (telescope). $\sum =\frac{1}{2}[g(1)-g(n+1)]=\frac{1}{2}(\frac{1}{1\cdot 2}-\frac{1}{(n+1)(n+2)}).$ Step 3 (simplify). ${\sum }_{k=1}^n\frac{1}{k(k+1)(k+2)}=\frac{1}{4}-\frac{1}{2(n+1)(n+2)}=\frac{n(n+3)}{4(n+1)(n+2)}.$ Check $n=3$: direct $=\frac{1}{6}+\frac{1}{24}+\frac{1}{60}=\frac{20+5+2}{120}=\frac{27}{120}=\frac{9}{40}$; formula $=\frac{3\cdot 6}{4\cdot 4\cdot 5}=\frac{18}{80}=\frac{9}{40}$. ✓ So the answer is $\frac{9}{40}$ at $n=3$.

Step 1 (exact partial sum). From L4.1, $S_n=\sum _{k=1}^n\frac{1}{(2k-1)(2k+1)}=\frac{n}{2n+1}.$ WHY telescoping is the key: it hands us an exact closed form for the partial sum, so the infinite sum is just its limit (see Convergence of Series). Step 2 (limit). As $n\to \infty$, divide top and bottom by $n$: ${\lim }_{n\to \infty }\frac{n}{2n+1}={\lim }_{n\to \infty }\frac{1}{2+\frac{1}{n}}=\frac{1}{2}.$ Step 3 (conclusion). The partial sums approach $\frac{1}{2}$, so the series converges with sum $\frac{1}{2}$.

Step 1 (the identity that "knows about $k^2$"). Expand $(k+1{)}^3-k^3=3k^2+3k+1$. WHY this tool: summing a difference telescopes, and this particular difference contains $k^2$ — so summing it lets us solve for $\sum k^2$. Step 2 (sum both sides, left telescopes). ${\sum }_{k=1}^n[(k+1{)}^3-k^3]=(n+1{)}^3-1^3=(n+1{)}^3-1.$ ${\sum }_{k=1}^n(3k^2+3k+1)=3\sum k^2+3\cdot \frac{n(n+1)}{2}+n.$ Step 3 (solve for $\sum k^2$). Set them equal: $3\sum k^2=(n+1{)}^3-1-\frac{3n(n+1)}{2}-n.$ Expand $(n+1{)}^3-1=n^3+3n^2+3n$. Then $3\sum k^2=n^3+3n^2+3n-\frac{3n^2+3n}{2}-n=n^3+\frac{3n^2}{2}+\frac{n}{2}=\frac{2n^3+3n^2+n}{2}=\frac{n(n+1)(2n+1)}{2}.$ Step 4. Divide by $3$: $\sum _{k=1}^n{k}^2=\frac{n(n+1)(2n+1)}{6}.$ $■$ A rigorous alternative is Mathematical Induction — but the telescope derives the formula instead of merely verifying it.

Step 1 (split by linearity — two independent worlds). ${\sum }_{k=1}^n{k}^3+{\sum }_{k=1}^n\frac{1}{k(k+1)}.$ Step 2 (standard cube sum + telescope from L3.1). ${\sum }_{k=1}^n{k}^3={\left[\frac{n(n+1)}{2}\right]}^2,{\sum }_{k=1}^n\frac{1}{k(k+1)}=\frac{n}{n+1}.$ Step 3 (combine). ${\sum }_{k=1}^n[k^3+\frac{1}{k(k+1)}]={\left[\frac{n(n+1)}{2}\right]}^2+\frac{n}{n+1}.$ Evaluate $n=4$: cubes $=(\frac{4\cdot 5}{2}{)}^2=10^2=100$; telescope $=\frac{4}{5}$. Total $=100+\frac{4}{5}=\frac{504}{5}=100.8.$

L1 read & count → L2 split with linearity + standard sums → L3 spot telescoping via partial fractions → L4 design the split (mind the scale factor) → L5 prove, take limits, combine. Every telescope obeys the same heart: first survivor minus last survivor.