Complete tails for geometric and telescoping series

Goal and scope

Prove convergence of two series with exact partial sums and remainders.

Only full declared laws: 2^{-n} starting at zero and 1/[n(n+1)] starting at one.

Givens and symbols

• algebra
• epsilon
• definition_here

• G1 — S_N=sum from n=0 to N of 2^{-n}; T_N=sum from n=1 to N of 1/[n(n+1)], N≥1.

• S_N,T_N — two partial sums
• R_N — geometric remainder

Rules and sources

• mko_definition · cited — This slice handles exact geometric and telescoping series. Retain the starting-index convention, partial sum and remainder. The geometric sum formula is used only for |r|<1. A long finite prefix or terms tending to zero cannot alone establish convergence. · mko-infinite-series
• algebra · prerequisite — Use real-field identities; division requires a nonzero denominator and multiplication by a positive number preserves inequalities.
• epsilon · prerequisite — Convergence requires for every positive epsilon an integer N controlling every n≥N by a strict error bound; use the Archimedean property of the reals.
• definition_here · prerequisite — An infinite-series sum is the limit of finite partial sums with the specified starting index.

Main derivation

S_N=\sum_{n=0}^{N}2^{-n}

S_N-\tfrac12S_N=1-2^{-(N+1)}

\tfrac12 S_N=1-2^{-(N+1)}

S_N=2-2^{-N}

R_N=2-S_N

R_N=2^{-N}>0

2^N\ge N+1\quad(N\ge0)

0<R_N\le1/(N+1)\to0

\frac1{n(n+1)}

\frac1n-\frac1{n+1}

T_N=\sum_{n=1}^{N}\left(\frac1n-\frac1{n+1}\right)

T_N=1-\frac1{N+1}

1-T_N=1/(N+1)

\forall\varepsilon>0\ \exists M\ \forall N\ge M:\ |T_N-1|<\varepsilon

\lim S_N=2,\quad\lim T_N=1

\sum_{n=0}^{\infty}2^{-n}=2,\quad\sum_{n=1}^{\infty}\frac1{n(n+1)}=1

Substeps and lemmas

No substeps in this guide.

Reverse audits

• S08 · premise_audit · S04, S07 — No convergent geometric formula is used for |r|≥1.
• S07 · premise_audit · S06 — Terms tending to zero would not suffice; here the partial-sum limit is established.
• S06 · inverse_equivalence · G1, S05 — The final negative term -1/(N+1) must remain.
• S05 · inverse_equivalence · G1 — Index zero is excluded from this series.
• S04 · premise_audit · S03 — This is an all-N induction and all-tail epsilon bound, not a large-N sample.
• S03 · inverse_equivalence · S02 — The remainder is nonzero for finite N.
• S02 · inverse_equivalence · S01 — The nonzero multiplier makes this reversible.
• S01 · inverse_equivalence · G1 — The remaining last exponent is N+1, not N.

Error controls and negative controls

Starting indices

The geometric law starts at zero and the telescoping law at one.

Partial sums versus limits

Give the exact remainder before proving it tends to zero.

• Finite S_N already equals 2. → reject — The remainder 2^{-N} is positive.
• Terms tending to zero force convergence. → reject — Harmonic blocks each contribute at least 1/2 and diverge.
• At r=1 the geometric series still sums to 1/(1-r). → reject — The denominator vanishes and partial sums N+1 diverge.

Result and limitations

The sums are 2 and 1, with exact tail errors 2^{-N} and 1/(N+1).

Verification and source binding

2 declared checks for this guide; only the listed expressions are covered.

• source commit: b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f
• adopted source SHA-256: 25a2460da5141a254a90cac613296ae05ae74dabff88232cf2d5997acececa66
• original source SHA-256: 25a2460da5141a254a90cac613296ae05ae74dabff88232cf2d5997acececa66
• Lean: NOT_RUN · independent AI: SCOPED_LEAD_AND_BILINGUAL_CONTEXT_REVIEW

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