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Part BCSIR NET June 2023residue-by-series-product

Residue by series product

Let f(z)=exp⁡(z+1/z),z∈Cf(z) = \exp(z + 1/z), z \in \mathbb{C}∖{0}. The residue of f at z = 0 is

  1. A.∑l=0∞\sum_{l=0}^\infty 1/(l+1)!
  2. B.∑l=0∞\sum_{l=0}^\infty 1/(l!(l+1))
  3. C.∑l=0∞\sum_{l=0}^\infty 1/(l!(l+1)!)✓ Correct
  4. D.∑l=0∞\sum_{l=0}^\infty 1/((l2+l)!)1/((l^{2}+l)!)

The idea, in plain words

Multiply the two exponential series, one in the variable and one in its reciprocal, and collect everything that lands on the single reciprocal power. Each such contribution pairs one term from each series, so the answer is an infinite sum of reciprocal factorial products.

Solution

Multiply the series eᶻ =∑zm/m= \sum z^{m}/m! and e1/ze^{1/z} =∑= \sum z−k/kz^{-k}/k!; the coefficient of z−1z^{-1} comes from k=m+1:∑m1/(m!(m+1)!)k = m + 1: \sum_m 1/(m!(m+1)!).

Why each option is right or wrong

Checked against 3 symbolic computations
  • A.WrongExecution slip

    False — only one factorial appears

    so this is the series for a single exponential rather than a product of two.

    SymPy's partial sums separate them clearly: the first four terms of this series give 41/24 ≈ 1.708

    while the correct series gives 229/144 ≈ 1.590.

  • B.Wrong

    False.

    The second factor is (l + 1), not (l + 1)!.

    This is what you get by tracking the index shift but forgetting that the coefficient of z−kz^{-k} in e1/ze^{1/z} is 1/k!, not 1/k.

  • C.Correct

    True.

    Multiply the two Laurent series: eᶻ =∑mzm/m= \sum_{m} z^{m}/m! and e1/ze^{1/z} =∑k= \sum_{k} z−k/kz^{-k}/k!.

    A term contributes to z−1z^{-1} exactly when m − k = −1, that is k = m + 1 — SymPy confirms that index shift.

    Collecting those terms gives ∑m1/(m!(m+1)!)\sum_{m} 1/(m!(m+1)!), whose partial sums SymPy evaluates as 229/144 ≈ 1.590 after four terms (the exact value is the Bessel function I1(2))I_{1}(2)).

    The trigger: for a product of two known series, write down the index equation that produces the power you want, then sum over the free index.

  • D.WrongExecution slip

    False.

    (l2+l)(l^{2} + l)! is a single factorial of a product-like expression

    which never arises from multiplying two series — the two factorials stay separate.

    It is the right shape only if you mistakenly combine m! and (m+1)! into one factorial of m(m+1).

2 of these 3 wrong options correspond to a named reasoning error — the same errors recur across subjects.

The trap it tests

Execution slip

The idea was right. The computation was not.

Drill statements like this

Related counterexample: |f| bounded near an isolated singularity ⇒ pole

More on this topic

The chapter behind this: Singularities — three types, three tests — free to read

From Singularities and Residues › Laurent series, classification of singularities, Casorati–Weierstrass

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

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