RAYAT SHIKSHAN SANSTHA’S S.M.JOSHI COLLEGE HADAPSAR, PUNE

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RAYAT SHIKSHAN SANSTHA’S S. M. JOSHI COLLEGE HADAPSAR, PUNE
RAYAT SHIKSHAN SANSTHA’S S.M.JOSHI COLLEGE, HADAPSAR, PUNE
RAYAT SHIKSHAN SANSTHA’S S.M.JOSHI COLLEGE HADAPSAR, PUNE
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RAYAT SHIKSHAN SANSTHA’S S.M.JOSHI COLLEGE HADAPSAR, PUNE
RAYAT SHIKSHAN SANSTHA’S S.M.JOSHI COLLEGE HADAPSAR, PUNE
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RAYAT SHIKSHAN SANSTHA’S S.M.JOSHI COLLEGE HADAPSAR, PUNE
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RAYAT SHIKSHAN SANSTHA’S S.M.JOSHI COLLEGE HADAPSAR, PUNE
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SEQUENCES AND SERIES.
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RAYAT SHIKSHAN SANSTHA’S S.M.JOSHI COLLEGE HADAPSAR, PUNE-411028 PRESANTATION BY Prof . DESAI S.S Mathematics department Subject – Complex Analysis Topic – Sequence ,Series and their Convergence

Sequences Sequence For example, the sequence {1 + in} is (1) If limnzn = L, we say this sequence is convergent. See Fig bellow. Definition of the existence of the limit:

Fig : Illustration

Example 1 The sequence converges, since See Fig 19.2.

******* THEOREM A sequence {zn} converge to a complex number L if and only if Re(zn) converges to Re(L) and Im(zn) converges to Im(L). Criterion for Convergence

Example 2 The sequence converges to i. Note that Re(i) = 0 and Im(i) = 1. Then

Series An infinite series of complex numbers is convergent if the sequence of partial sum {Sn}, where converges.

Geometric Series For the geometric Series (2) the nth term of the sequence of partial sums is (3) and (4)

****** Since zn  0 as n   whenever |z| < 1, we conclude that (2) converges to a/(1 – z) when |z| < 1 ; the series diverges when |z|  1. The special series (5) (6) valid for |z| < 1.

Example 3 The series is a geometric series with a = (1 + 2i)/5 and z = (1 + 2i)/5. Since |z| < 1, we have

******* If converges, then If then the series diverges. THEOREM If converges, then Necessary Condition for Convergence If then the series diverges. THEOREM The nth Term Test for Divergence

***** An infinite series is said be absolutely DEFINITION An infinite series is said be absolutely convergent if converges. Absolute Convergence

Example 4 The series is absolutely convergent since |ik/k2| = 1/k2 and the real series converges. As in real calculus, Absolute convergence implies convergence. Thus the series in Example 4 converges.

****** THEOREM Suppose is a series of nonzero complex terms such that (9) (i) If L < 1, then the series converges absolutely. (ii) If L > 1 or L = , then the series diverges. (iii) If L = 1, the test is inconclusive. Ratio Test

******* THEOREM Suppose is a series of complex terms such that (10) (i) If L < 1, then the series converges absolutely. (ii) If L > 1 or L = , then the series diverges. (iii) If L = 1, the test is inconclusive. Root Test

Power Series An infinite series of the form (11) where ak are complex constants is called a power series in z – z0. (11) is said to be centered at z0 and z0 is referred to the center of the series.

Circle of Convergence Every complex power series has radius of convergence R and has a circle of convergence defined by |z – z0| = R, 0 < R < . See Fig 19.3.

The radius R can be (i) zero (converges at only z = z0) The radius R can be (i) zero (converges at only z = z0). (ii) a finite number (converges at all interior points of the circle |z − z0| = R). (iii)  (converges for all z). A power series may converge at some, all, or none of the points on the circle of convergence.

Example 5 Consider the series , by ratio test Thus the series converges absolutely for |z| < 1 and the radius of convergence R = 1.

THANK YOU