Cauchy product of series

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Prove that ∑n=0∞1n!⋅∑n=0∞(−1)nn!=1\displaystyle \sum_{n=0}^{\infty } \frac{1}{n!} \cdot \sum_{n=0}^{\infty } \frac{(-1)^{n}}{n!}=1.

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Prove that (∑n=0∞qn)2=∑n=0∞(n+1)qn(∣q∣<1)\displaystyle \left(\sum_{n=0}^{\infty } q^{n}\right)^{2}=\sum_{n=0}^{\infty } (n+1)q^{n} \quad (\left|q\right|<1).

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Prove that the square of the convergent series ∑n=1∞(−1)n+1n\displaystyle \sum_{n=1}^{\infty } \frac{(-1)^{n+1}}{\sqrt{n}} is a divergent series.

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Prove that the product of the two series ∑n=1∞(−1)n−1nα (α>0)\displaystyle \sum_{n=1}^{\infty } \frac{(-1)^{n-1}}{n^{\alpha }} \ (\alpha >0) and ∑n=1∞(−1)n−1nβ (β>0)\displaystyle \sum_{n=1}^{\infty } \frac{(-1)^{n-1}}{n^{\beta }} \ (\beta >0) is a convergent series when α+β>1\alpha +\beta >1, and a divergent series when α+β<1\alpha +\beta <1.

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Verify that the product of the two divergent series 1−∑n=1∞(32)n1-\displaystyle \sum_{n=1}^{\infty } \left(\frac{3}{2}\right)^{n} and 1+∑n=1∞(32)n−1(2n+12n+1)1+\displaystyle \sum_{n=1}^{\infty } \left(\frac{3}{2}\right)^{n-1}\left(2^{n}+\frac{1}{2^{n+1}}\right) is an absolutely convergent series.

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Let ∑n=1∞cn\sum_{n = 1}^{\infty } c_{n} be the series obtained by multiplying the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} and ∑n=1∞bn\sum_{n = 1}^{\infty } b_{n}. Find cnc_{n}, if:

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(1)

an=an−1,bn=bn−1,∣a∣<1,∣b∣<1a_{n} = a^{n-1}, b_{n} = b^{n-1},\left|a\right| < 1,\left|b\right| < 1;

(2)

an=qn−1,bn=qnn(n+1),∣q∣<1a_{n} = q^{n-1}, b_{n} = \frac{q^{n}}{n(n+1)},\left|q\right| < 1;

(3)

an=2n−1(n−1)!,bn=3n−1(n−1)!a_{n} = \frac{2^{n-1}}{(n-1) !}, b_{n} = \frac{3^{n-1}}{(n-1) !};

(4)

an=3n−1(n−1)!,bn=(−1)n−1(n−1)!a_{n} = \frac{3^{n-1}}{(n-1) !}, b_{n} = \frac{(-1)^{n-1}}{(n-1) !}

(5)

an=(−1)n(2n−1)!,bn=(−1)n−1(2n−2)!a_{n} = \frac{(-1)^{n}}{(2 n-1) !}, b_{n} = \frac{(-1)^{n-1}}{(2 n-2) !}.

Show that the square of the convergent series ∑n=1∞(−1)nn\sum_{n = 1}^{\infty } \frac{(-1)^{n}}{\sqrt{n}} is a divergent series.

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Let α>0,β>0\alpha > 0, \beta > 0. Show that the product of two convergent series

∑n=1∞(−1)nnα and ∑n=1∞(−1)nnβ \sum _{n = 1}^{\infty } \frac{(-1)^{n}}{n^{\alpha }} \quad \text{ and } \quad \sum _{n = 1}^{\infty } \frac{(-1)^{n}}{n^{\beta }}

is a series that converges if α+β>1\alpha +\beta > 1, and diverges if α+β<1\alpha +\beta < 1.

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Show that:

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(1)

(∑n=0∞ann!)(∑n=0∞(−a)nn!)=1\left(\sum_{n = 0}^{\infty } \frac{a^{n}}{n !}\right)\left(\sum_{n = 0}^{\infty } \frac{(-a)^{n}}{n !}\right) = 1;

(2)

(∑n=0∞ann!)2=∑n=0∞(2a)nn!\left(\sum_{n = 0}^{\infty } \frac{a^{n}}{n !}\right)^{2} = \sum_{n = 0}^{\infty } \frac{(2 a)^{n}}{n !}.

Show that the series obtained by multiplying two divergent series

1−∑n=1∞(32)n and 1+∑n=1∞(32)n−1(2n+2−(n+1)) 1-\sum _{n = 1}^{\infty }\left(\frac{3}{2}\right)^{n} \quad \text{ and } \quad 1+\sum _{n = 1}^{\infty }\left(\frac{3}{2}\right)^{n-1}\left(2^{n}+2^{-(n+1)}\right)

converges absolutely.

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Let the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n}, where an>0(n∈N)a_{n} > 0(n \in N), converge, and let the series ∑n=1∞bn\sum_{n = 1}^{\infty } b_{n}, where bn>0(n∈N)b_{n} > 0(n \in N), diverge. Show that the product of these series is a divergent series.

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Prove that if the series

∑n=1∞an,∑n=1∞bn and ∑n=1∞(∑k=1nakbn−k+1) \sum _{n = 1}^{\infty } a_{n}, \quad \sum _{n = 1}^{\infty } b_{n} \quad \text{ and } \quad \sum _{n = 1}^{\infty }\left(\sum _{k = 1}^{n} a_{k} b_{n-k+1}\right)

converge and their sums equal A,BA, B, and CC, respectively, then the equality C=ABC = A B holds.

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Prove that if the series ∑n=0∞an\sum_{n = 0}^{\infty } a_{n} and ∑n=0∞bn\sum_{n = 0}^{\infty } b_{n} are summable by the Poisson-Abel method to the sums AA and BB, respectively, then the series ∑n=0∞(a0bn+a1bn−1+…+an−1bn+anb0)\sum_{n = 0}^{\infty }\left(a_{0} b_{n}+a_{1} b_{n-1}+\ldots +a_{n-1} b_{n}+a_{n} b_{0}\right), i.e., the product of the given series, is also summable by the Poisson-Abel method to the sum ABA B.

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What conditions must the series u0+u1+u2+⋯+un+⋯u_{0} + u_{1} + u_{2} + \cdots + u_{n} + \cdots satisfy in order that its Cauchy product [Problem 34 of Part I, Problem 23 of Part II, Part VIII Chapter 1 5] with any convergent series v0+v1+v2+⋯+vn+⋯v_{0} + v_{1} + v_{2} + \cdots + v_{n} + \cdots results in a convergent series

u0v0+(u0v1+u1v0)+(u0v2+u1v1+u2v0)+⋯+(u0vn+u1vn−1+⋯+un−1v1+unv0)+⋯? u_{0}v_{0} + (u_{0}v_{1}+u_{1}v_{0}) + (u_{0}v_{2}+u_{1}v_{1}+u_{2}v_{0}) + \cdots + (u_{0}v_{n}+u_{1}v_{n-1}+\cdots +u_{n-1}v_{1}+u_{n}v_{0}) + \cdots ?
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Prove that the Cauchy product of two absolutely convergent series converges absolutely.

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References

[Dem97] Demidovich B.P. Сборник задач и упражнений по математическому анализу. 13-е изд., Издательство Московского университета / Издательство "ЧеРо", 1997.
[ККЧ+03] Кудрявцев Л.Д., Кутасов А.Д., Чехлов В.И., Шабунин М.И. Интегралы. Ряды. 2-е изд., ФИЗМАТЛИТ, 2003.
[PóSz98] Pólya G., Szegő G. Problems and Theorems in Analysis. Springer, 1998.
[Rud76] Rudin W. Principles of Mathematical Analysis. 3rd ed., McGraw-Hill, 1976.