Sandwiching the partial sum between two integrals
Fix a positive, decreasing function f with
a_n = f(n). The whole proof is reading off two rectangle pictures
against the curve, then squeezing.
Step 1 — over-estimate the area with left rectangles. On each interval
[k, k+1] the function is decreasing, so its biggest value is at
the left end, f(k) = a_k. A rectangle of height
a_k and width 1 therefore covers
the area under the curve on that strip:
a_k \;\ge\; \int_{k}^{k+1} f(x)\,dx.
Step 2 — sum the left rectangles from 1 to
n. The integrals join up end to end into one big integral:
\sum_{k=1}^{n} a_k \;\ge\; \sum_{k=1}^{n} \int_{k}^{k+1} f(x)\,dx = \int_{1}^{n+1} f(x)\,dx.
Step 3 — under-estimate the area with right rectangles. Now the smallest
value on [k-1, k] is at the right end,
f(k) = a_k, so that rectangle fits underneath the curve:
a_k \;\le\; \int_{k-1}^{k} f(x)\,dx.
Step 4 — sum the right rectangles from 2 to
n (the first term a_1 has no
strip to its left, so set it aside):
\sum_{k=2}^{n} a_k \;\le\; \int_{1}^{n} f(x)\,dx, \qquad\text{so}\qquad \sum_{k=1}^{n} a_k \;\le\; a_1 + \int_{1}^{n} f(x)\,dx.
Step 5 — the sandwich. Stacking Steps 2 and 4 traps the partial sum
s_n = \sum_{k=1}^{n} a_k between two integrals:
\int_{1}^{n+1} f(x)\,dx \;\le\; s_n \;\le\; a_1 + \int_{1}^{n} f(x)\,dx.
Step 6 — read off both directions. If the integral
\int_1^\infty f converges, the right-hand bound stays finite, so
the increasing partial sums s_n are bounded above and must
converge. If the integral diverges, the left-hand bound
\int_1^{n+1} f \to \infty drags s_n to
infinity with it. Either way:
\sum_{n=1}^{\infty} a_n \text{ converges} \;\iff\; \int_{1}^{\infty} f(x)\,dx \text{ converges.}
Let f be positive, continuous and decreasing on
[1, \infty), and set a_n = f(n). Then:
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Same fate. \sum_{n=1}^{\infty} a_n converges
if and only if \int_{1}^{\infty} f(x)\,dx converges.
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Sandwich bound. For every n,
\int_{1}^{n+1} f \;\le\; \sum_{k=1}^{n} a_k \;\le\; a_1 + \int_{1}^{n} f.
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The p-series. Applying the test to
f(x) = 1/x^{p},
\sum_{n=1}^{\infty} \frac{1}{n^{p}} \text{ converges} \iff p > 1.
In particular the harmonic series (p = 1) diverges, while
\sum 1/n^2 converges.
The p-series, settled
Apply the test to f(x) = x^{-p} (positive and decreasing for
p > 0). For p \ne 1,
\int_{1}^{\infty} x^{-p}\,dx = \left[\frac{x^{1-p}}{1-p}\right]_{1}^{\infty}.
The exponent 1 - p decides everything: if
p > 1 then 1 - p < 0 and
x^{1-p} \to 0, leaving the finite value
\tfrac{1}{p-1}; if p < 1 the exponent is
positive and the integral blows up. The boundary case
p = 1 gives \int_1^\infty x^{-1}\,dx = [\ln x]_1^\infty = \infty.
Hence convergence exactly when p > 1.
The same rectangle picture does more than decide convergence — it bounds the tail.
The remainder after n terms,
R_n = \sum_{k=n+1}^{\infty} a_k, is itself sandwiched by
integrals of the leftover region:
\int_{n+1}^{\infty} f(x)\,dx \;\le\; R_n \;\le\; \int_{n}^{\infty} f(x)\,dx.
For \sum 1/n^2, this gives
R_n \le \int_n^\infty x^{-2}\,dx = \tfrac1n: to pin the sum to
three decimals you need only about a thousand terms, and you know it
without summing the infinite tail. A test that decides convergence usually tells
you, for free, how fast you are converging.