Visual walkthrough — Telescoping series

We build everything from zero. If a symbol appears, it was drawn first.

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Step 1 — What is a "series" anyway? A running total

WHAT. A series is what you get when you keep a running total of a list of numbers. Call the numbers $a_1,a_2,a_3,\dots$ — just "term number 1, term number 2, ..." The little number below is the index: it tells you which term you are pointing at, nothing more.

The partial sum $S_N$ means: "stop after $N$ terms and tell me the total so far."

$$S_N=\underset{\text{term 1}}{\underbrace{a_1}}+\underset{\text{term 2}}{\underbrace{a_2}}+\cdots +\underset{\text{term }N}{\underbrace{a_N}}$$

WHY. Infinity is scary; a running total is not. We will never add infinitely many things. We add $N$ of them, get a clean answer, and only at the very end ask "what does that answer drift toward as $N$ grows?"

PICTURE. Each term is a brick. Stacking bricks left to right, the height of the stack after $N$ bricks is $S_N$.

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Step 2 — The special ingredient: every term is a difference

WHAT. A telescoping series is one where each brick $a_n$ is secretly a subtraction of two heights drawn from a single background sequence $b_1,b_2,b_3,\dots$:

$$a_n=\underset{\text{my own height}}{\underbrace{b_n}}-\underset{\text{the next one’s height}}{\underbrace{b_{n+1}}}$$

Here $b_n$ is just some sequence — think of it as a fence with posts of decreasing height. Term $a_n$ is the drop from post $n$ to post $n+1$.

WHY. Differences are magic for adding because a drop shares an edge with the next drop. Post $n+1$ is the bottom of drop $n$ and the top of drop $n+1$. That shared post is what will cancel.

PICTURE. Below: the fence of posts $b_n$ (black), and each term $a_n$ shown as the red vertical gap between two neighbouring posts.

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Step 3 — Add the first few, and watch a post get used twice

WHAT. Add just the first three terms, writing each one out fully instead of collapsing it:

$$S_3=(\stackrel{+}{\overbrace{b_1}}-\stackrel{-}{\overbrace{b_2}})+(\stackrel{+}{\overbrace{b_2}}-\stackrel{-}{\overbrace{b_3}})+(\stackrel{+}{\overbrace{b_3}}-\stackrel{-}{\overbrace{b_4}})$$

Look at $b_2$: it appears once with a minus (end of term 1) and once with a plus (start of term 2). Same for $b_3$.

WHY. This is the whole trick, caught in the act. A post in the middle of the fence is the bottom of one drop and the top of the next. Those two roles carry opposite signs, so the post is counted $+b_2$ and $-b_2$ — a net of zero.

PICTURE. The red posts $b_2,b_3$ each get a $+$ tag from one term and a $-$ tag from its neighbour. Their tags stare at each other, ready to annihilate.

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Step 4 — The cancellation: interior posts vanish

WHAT. Group the plus/minus pairs:

$$S_3=b_1+\underset{=0}{\underbrace{(-b_2+b_2)}}+\underset{=0}{\underbrace{(-b_3+b_3)}}-b_4$$

Every interior post cancels perfectly, leaving

$$S_3=b_1-b_4$$

WHY. Only $b_1$ (nothing before it to supply a $+$/$-$ partner) and $-b_4$ (nothing after it) are lonely. Everybody in between found their opposite twin.

PICTURE. The middle posts fade to grey (cancelled); only the first post $b_1$ and the last post $b_4$ glow red — the survivors.

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Step 5 — General $N$: first minus last

WHAT. Nothing about "$3$" was special. For any $N$:

$$S_N=(b_1-b_2)+(b_2-b_3)+\cdots +(b_N-b_{N+1})=b_1-b_{N+1}$$

Each term $b_k$ for $2\le k\le N$ shows up once as $+b_k$ and once as $-b_k$. Gone. Survivors: the very first $b_1$ and the very last $-b_{N+1}$.

WHY. We generalise so we never have to write out the middle again. The bookkeeping is always the same: first minus last-plus-one.

PICTURE. A long fence. A single red arrow spans from post $b_1$ straight down to post $b_{N+1}$ — the whole sum is just that one net gap, no matter how many posts sit between them.

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Step 6 — The infinite sum: let the last post walk off to the horizon

WHAT. To sum forever, let $N\to \infty$. The lead survivor $b_1$ is frozen; only the trailing survivor $b_{N+1}$ moves:

$$\sum _{n=1}^{\infty }(b_n-b_{n+1})=\underset{N\to \infty }{\lim }{S}_N=b_1-\underset{\text{call it }L}{\underbrace{\underset{N\to \infty }{\lim }{b}_{N+1}}}$$

WHY. We use a limit because we can't reach the end of an infinite fence — we can only ask what height the last post approaches as it marches to the horizon (see Limits of Sequences). Whatever that height $L$ is, the answer is $b_1-L$.

PICTURE. The trailing post slides rightward toward a dashed horizon line at height $L$. The red gap the answer measures settles from $b_1-b_{N+1}$ down to the final $b_1-L$.

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Step 7 — Degenerate case: what if the last post never settles?

WHAT. The formula $\text{sum}=b_1-L$ only makes sense if $L$ exists (is a single finite number). If $b_n$ grows without bound, or bounces forever, there is no $L$ — and the series diverges (no exact sum).

WHY. We must cover this or the reader will trust the collapse blindly. Example: $b_n=n$, so $a_n=n-(n+1)=-1$. Then $S_N=b_1-b_{N+1}=1-(N+1)=-N$. As $N$ grows, $S_N\to -\infty$. The telescope collapsed correctly — but the trailing post ran off to infinity, so there is no finite total.

PICTURE. Two fences side by side: on the left posts sink to a flat horizon ($L$ exists → converges); on the right posts climb forever (no $L$ → diverges). The red trailing arrow on the right just keeps stretching.

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The one-picture summary

Everything above, compressed: a fence of posts $b_1,b_2,\dots$; each term is the red drop between neighbours; all interior posts pair off and vanish; only $b_1$ and the runaway $b_{N+1}$ (heading to $L$) remain — so the whole infinite sum is the single net gap $b_1-L$.

Recall Feynman retelling — the whole walkthrough in kid-words

Picture a long fence made of posts, each a bit shorter than the last. A "term" isn't a post — it's the little step down from one post to the next. Now I want to add up all the steps. Here's the beautiful part: the bottom of one step is the exact same spot as the top of the next step. So when I add "step down from post 2" and then "step up would-be from post 2," that shared post gets counted once as plus and once as minus — poof, it cancels. Every post in the middle of the fence is shared by two steps, so every middle post cancels. Only two posts are lonely: the very first one (no step before it) and the very last one (no step after it). So the total of ALL the steps is just: height of the first post minus height of the last post. If the fence goes on forever and the posts sink toward some flat ground at height $L$, the "last post" becomes that flat ground — and the whole endless sum is simply first post minus $L$. That's a telescope folding shut: the middle slides away, only the two rims are left.

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Connections

• Telescoping series (Hinglish) — the parent topic this page zooms into.
• Partial Fraction Decomposition — how you manufacture the $b_n-b_{n+1}$ shape.
• Sequence of Partial Sums — the list $S_N$ we watched collapse.
• Limits of Sequences — decides where the trailing post lands.
• Convergence Tests for Series — telescoping beats these by giving the exact value.
• Method of Differences — the same cancellation idea for finite sums.
• Geometric Series — the other family with an exact sum; good contrast.