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### Sum of 1/Sqrt(i)

```
Date: 11/20/2000 at 13:11:23
From: Jeff Khoury
Subject: Mathematical induction, sum

What's the formula for

n
SUM [1/sqrt(i)]
i=1

This question was asked in a college quiz.
```

```
Date: 11/20/2000 at 16:21:17
From: Doctor Rob
Subject: Re: Mathematical induction, sum

Thanks for writing to Ask Dr. Math, Jeff.

I don't know a formula for the sum. I do know that the sum lies
between 2*sqrt(n)-2 and 2*sqrt(n). That is because what you have is
bounded above by

INT [1/sqrt(t) dt] from 0 to n

and below by

INT [1/sqrt(t) dt] from 1 to n

which have the values I indicated.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 11/21/2000 at 11:59:32
From: Jeff
Subject: Re: Mathematical induction, sum

Sorry to bother you again, but if my math teacher asked for a formula
then it should exist. How come you can't find it? Thank you.
```

```
Date: 11/21/2000 at 12:57:55
From: Doctor Rob
Subject: Re: Mathematical induction, sum

Thanks for writing again.

for two days, and asked several very knowledgeable friends. We all
believe that no such formula exists. If your instructor has such a
formula, we would be very interested in seeing it.

I do know that the above sum diverges as n increases without bound,
because of the limits I quoted above. In fact, my friends and I
believe that

n
lim     [SUM 1/sqrt(i) - 2*sqrt(n)]
n->infinity  i=1

is a well-known quantity, zeta(1/2) = -1.4603545..., which is believed
to be a transcendental number. Of course this limit doesn't give you a
formula for the sum.

Sorry!

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 11/21/2000 at 15:54:56
From: Doctor Rob
Subject: Re: Mathematical induction, sum

Dear Jeff,

One of my friends has devised a formula for your sum. It is not
simple, but here it is:

n
SUM k^(-1/2) = 2*n^(1/2) - 3/2 + n^(-1/2)/2 -
k=1
n
(1/2)*INT (x-[x]-1/2)*x^(-3/2) dx)
1

Here [x] means the greatest integer less than x. This seems very
likely to be amenable to proof by induction.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 11/22/2000 at 08:49:55
From: Jeff
Subject: Re: Mathematical induction, sum

Dear Dr. Rob,

I have to tried to prove the formula you gave me by induction and
failed. I'm asking you (I hope I'm not asking too much) to send me the
proof of this formula.

Thank you a lot for your help.
```

```
Date: 11/27/2000 at 08:58:44
From: Doctor Rob
Subject: Re: Mathematical induction, sum

Thanks for writing back, Jeff.

First you need the basis of the induction. When n = 1, both sides are
equal to zero, because the integral is from 1 to 1, and hence has
value zero.

Now assume it is true for n, and try to prove it for n+1. Then
starting with the right side for n+1, you have

2*(n+1)^(1/2) - 3/2 + (n+1)^(-1/2)/2 -

n+1
(1/2)*INT [x-[x]-1/2)*x^(-3/2) dx
1

= 2*(n+1)^(1/2) - 3/2 + (n+1)^(-1/2)/2 -

n                  n+1
(1/2)*INT f(x) dx + (1/2)*INT f(x) dx
1                   n

Now use the induction hypothesis to replace the first integral by its
equal from the statement for n, which you are assuming is true. The
last part is to evaluate the integral

n+1
INT (x-[x]-1/2)*x^(-3/2) dx
n

This is made easier since [x] = n in this interval of integration, so
you get

n+1
INT x^(-1/2) - n*x^(-3/2) - (1/2)*x^(-3/2) dx
n

which involves only integrating powers of x. When you do this, you
should get terms which cancel with the ones in the rest of the
equation, and you should be left with

n                            n+1
SUM k^(-1/2) + (n+1)^(-1/2) = SUM k^(-1/2)
k=1                           k=1

which is the left-hand side of the statement for n+1. You can see that
this proves that the right-hand and left-hand sides of the statement
for n+1 are equal, assuming they are for n, so the induction step is
complete.

Now invoke the Principle of Mathematical Induction to conclude that
the statement is true for all natural numbers n.

You fill in the details.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Number Theory
High School Number Theory
High School Sequences, Series

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