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Example. The series in which the ratio of the (n + 1)th to

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the nth term is 1 - is convergent, if x> 1.

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Unti/un 1 1 n (1 Un+1/un)

1 which is always less than a quantity - which is less than 1.

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453. (V.) If u, + U. + Uz +

and
V + V2 + V3 +

in each of which all the terms have the same sign be two series such that the limit of un: Vn and of vn:u, is finite, then both series are convergent or both series are divergent.

For since the limit of Un : vn is finite, the value of Un : Vn must always be less than some finite quantity k say. Thus un < kv, ; Uz <kv,, &c.

.. U1 + U2 + ... + Un +.. <k(0, + V2 + + 2 + ...). .. If the v-series is finite, the u-series is finite;

if the u-series is infinite, the v-series is infinite. In the same way, since the limit of vn : Un is finite, we may show that

If the u-series is finite, the v-series is finite;
if the v-series is infinite, the u-series is infinite.

454. (VI.) If f(n) always decreases as n increases and is always positive, then according as the series whose general term is f(n) is convergent or divergent so also is the series whose general

f (login) term is

where b is greater than 1.

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Since b> 1, .. as n increases, log, n increases, and :: f(log, n) decreases. (1) Take 6 to be integral.

f(logo b») i f(logo 6n+1) Consider the terms from

to

excluding

71+1 the last. The number of these terms is 6n+1 6.

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b

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n

and

f(log, 6n+1) > (61+1 6)

i.e. > (1 – 1/6) f (n + 1). bu+1

f(log, n) Thus, the ratio of the series

to the series f(n) lies between 1-1/b and b - 1. Hence the two series are either both convergent or both divergent.

(2) Take b to be fractional, and to lie between the two integers i and j. Then logo n lies between login and log; n.

f (logo n) Hence, for all values of b > 1, the series

is convergent or divergent according as the series f(n) is convergent or divergent.

455. Now the series 2 is convergent or divergent according as is less or not less than 1.

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is convergent or divergent according as x is less or not less than 1: i.e. according as logo x is less or not less than 0.

Thus the series nie is convergent or divergent according as x is less or not less than – 1.

The case of divergency here was proved more directly in Art. 451.

Cor. Hence all the series

(log njo {log (log n)} [log {log (log n}}]®, &c. &c.,
() [()}*

. n log n n log n log (log n) are convergent or divergent according as x is less or not less than -1.

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456. By means of the comparative theorem V, we may determine the character of

many

series. Thus if the nth term of a series can be expressed as a finite or convergent sum of terms involving descending powers of n, the series will be convergent or divergent according as the highest index of n is or not less than -1.

For the nth term ana + bny + where the indices are descending has a finite ratio to the nth term nie.

457. It should be noticed that the assigned characteristics in the above theorems may be supposed to begin in a given series at any term at a finite distance after the first. Thus U, in the above theorems means that term at which the assigned characteristic begins to hold'.

Principle of Equating Coefficients.

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458. If an endless series in powers of x is equal to zero, for every value of x from 0 to some finite value f, then each coefficient must separately be zero.

For let do, the coefficient of ac", be the first coefficient which is supposed not to be zero. Then, for the values of x in question,

QodX" + a22"+1 + A2X^+2 +

i.e. ac®+1 (Q4 + A.92 + .) QoX". If then x is not equal to zero, dividing by x",

2 (az + a+ ...) Qo, a finite constant, .. when w=f, the series within the bracket is finite; and it cannot become infinite by decreasing x.

But, if a, is not zero, this series would have to become greater than any assignable quantity when a becomes sufficiently small. But this has been shown to be impossible. Hence ao must be zero; and, therefore, every coefficient must be zero.

459. Hence, if two endless series in powers of x are equal to one another for every value of x between 0 and f, then the corresponding coefficients in the two series must be severally equal.

In the above it is essential to note that the two series must be convergent, otherwise we cannot assign any value to their sum, nor therefore speak of them as being equal.

The proof follows immediately from the preceding article. For, If

(lo + % + AnX? + = bo + b 2 + bxX2 + ... then

a, bo + (az – bı) x + (a, b.) x2 + 0. :. by above,

(to bo = a; - 6, = Az b 0,

ao = bo, a, = 61, a. = b

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Convergency of the Binomial Series. 460. Consider the series

m (m – 1)
am + mam-1x +

1.2
m (m – 1) ... (m - r + 1)

am-r 2" +

am-22:+ ...

+

r

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C

X

m +

(1-m+1)

1

r

a

a

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where m is finite.

m - p+1 Here the multiplier

Up We will take r to be algebraically > m + 1, so that the factor in the bracket is positive and the multiplier has the sign of x/a.

A. Let m +1 >0; then the multiplier is always arithmetically <x/a; but increases up to the limit xịa, as r increases.

Hence, if x/a is arithmetically <1, the multiplier is less than a quantity less than 1; .. the series is convergent.

If x/a is arithmetically > 1, then, when r is sufficiently large, the multiplier becomes greater than a quantity greater than 1; :. the series is divergent.

B. Let m +1 <0; then the multiplier is always arithmetically > x/a; but decreases down to the limit x/a, as r increases.

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m + 1

r

Hence, if x/a is arithmetically < 1, then, when r is sufficiently large, the multiplier becomes less than a quantity less than 1; .:. the series is convergent.

If x/a is arithmetically > 1, the multiplier is greater than a quantity greater than 1; .. the series is divergent.

Hence, if x/a is not + or -- 1, the series is convergent or divergent accordiny as x/a is arithmetically less or greater than 1.

461. Let x/a = -1; then the multiplier is positive when r is algebraically > m + 1, so that all the terms have the same sign, and the series is either convergent or divergent.

A. Let m + 1 <0; then the multiplier is greater than 1, and :. the series is divergent.

B. Let m +1>0; then the multiplier is less than 1, but increases up to the limit 1.

In this case
Ur-1/ur

1-(m + 1)/r 1 1
r (1 - U+1/ur)

m +1 :. if m + 1> 0 but < l, this expression, when r is sufficiently large, is greater than l; .. the series is divergent.

But if m +1>1, this expression is less than a quantity less than 1, .. the series is convergent.

Thus, when x/a=-1, the series is convergent or divergent according as m is positive or negative.

462. Let æ/a = +1; then the multiplier is negative when r is algebraically > m+ 1, so that the terms are alternately signed.

A. Let m +1>0; then the multiplier is arithmetically less than 1; .. the terms are arithmetically decreasing, and the series is finite for all values of n.

U,

r Moreover

m + 1

1+ 1 - m - 1

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m + 1
+
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r- m -

1 m

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m + 1
m + 1

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+

+
r - m
1 rm

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m + 1

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....(

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Unti

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