# Approximating series of fractions [duplicate]

Let $$P = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}}+ \frac{1}{\sqrt{4}} ... +\frac{1}{\sqrt{10000}}$$ what is the value of the floor function of P?

My try:

I tried assuming these 2 bounds

$$P_x = 1 + 1 + 1 + \frac{1}{2}+...\frac{1}{99 }$$ where it is repeated until the next square number (eg. there are 3 1's at the beginning of the sequence corresponding to the $$\frac{1}{\sqrt{2}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{4}}$$ where $$\frac{1}{\sqrt{4}}$$ is the next square number

and

$$P_y = \frac{1}{2} + \frac{1}{2} +\frac{1}{2} + \frac{1}{3} + \frac{1}{3} ... \frac{1}{100}$$ withe the same counting process as $$P_x$$

then we know that

$$P_x>P>P_y$$

$$99*2 + (\frac{1}{1} + \frac{1}{3} + \frac{1}{4} ... \frac{1}{99}) > P >99*2 -(\frac{1}{2} + \frac{1}{3} + \frac{1}{4} ... \frac{1}{100})$$

but as you can see, the floor function of P can be either 197 or 198, how would I answer this?

## marked as duplicate by Martin R, Paul Frost, Sil, rtybase, Robert Z sequences-and-series StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Mar 24 at 19:26

• Simple computer simulation: $197$... valuable for checking analytic solution. – David G. Stork Mar 24 at 7:16
• – Martin R Mar 24 at 7:37

You can use the following inequality. $$\frac{1}{\sqrt{k}}>2(\sqrt{k+1}-\sqrt{k}).$$ We obtain: $$\sum_{k=2}^{10000}\frac{1}{\sqrt{k}}>2\sum_{k=2}^{10000}(\sqrt{k+1}-\sqrt{k})=2(\sqrt{10001}-\sqrt2)>197.$$ Also, we have $$\frac{1}{\sqrt{k}}<2(\sqrt{k}-\sqrt{k-1}).$$

• what happened to the 2nd expression in the string ? if possible though, a solution without calculus would be nice – SuperMage1 Mar 24 at 7:13
• @SuperMage1 It was typo. I fixed. – Michael Rozenberg Mar 24 at 7:15

From the definition of the Riemann integral we can say:

$$\sum_\limits{n=2}^{10000} \frac {1}{\sqrt n} \le \int_1^{10000}\frac {1}{x^\frac 12}\ dx \le\sum_\limits{n=1}^{9999} \frac {1}{\sqrt n}$$

or

$$\int_2^{10001}\frac {1}{x^\frac 12}\ dx \le \sum_\limits{n=2}^{10000} \frac {1}{\sqrt n} \le \int_1^{10000}\frac {1}{x^\frac 12}\ dx$$

$$2 (\sqrt {10001} - \sqrt 2)\le\sum_\limits{n=2}^{1000} \frac {1}{\sqrt n} \le 2 (\sqrt {10000} - 1)$$

• It is Georg Friedrich Bernhard Riemann, not Reimann :) – Martin R Mar 24 at 7:42

Use $$\int_2^{10001}\frac{\mathrm{d}x}{\sqrt{x}}