Is it possible for Riemann Sum and Standard Integration to have different answers?

Question

I have a specific question for this equation :

$$\frac{1}{x+1} $$

Using Standard Integration,

$$\int_{0}^{1} \frac{1}{x+1} $$

which is approximately 0.69

Using Riemann Sum (right end point), however, I get this

$$ \lim _{n \to \infty } \Sigma ^n _{i=1} \frac{1}{n+i} $$

$$ = \lim _{n \to \infty } ( \frac {1} {n+1} + \frac {1} {n+2} + \frac {1} {n+3} + ... +\frac {1} {2n} ) $$

$$ = \lim _{n \to \infty } (\frac{1}{n})( \frac {1} {1+\frac{1}{n}} + \frac {1} {1+\frac{2}{n}} + \frac {1} {1+\frac{3}{n}} + ... +\frac {1} {2} ) $$

$$= (\frac{1}{\infty})( \frac {1} {1+0} + \frac {1} {1+0} + \frac {1} {1+0} + ... +\frac {1} {2} ) $$

$$= 0 $$

Is this right?

Answer 1

You cannot simply "plug in infinity" in the limit here and conclude the product is zero, because the number of terms in the sum is also increasing.

As to the broader question of under what conditions a Riemann sum can be computed by taking an antiderivative, see https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus#Second_part

Answer 2

To put this in a general sense, the task you are facing is evaluating the following type of limit: $$ \lim_{n\to\infty}\sum_{i=1}^nf(n,i), $$ in which $\lim_{n\to\infty}f(n,i)=0$ for any fixed $i$. However, this is not the case when $i$ is a dummy variable in the sum. Thus, you cannot simply plug in the pointwise limit of $f(n,i)$ to get zero.

Answer 3

For a function like this, it is not possible for the Riemann sum to be different to the integral.

Your second-to-last line is not correct, because the two terms you are multiplying together are:

1. $\frac{1}{n}$, which tends towards 0 as $n \rightarrow \infty$; and

2. $\frac{1}{1 + \frac{1}{n}} + \frac{1}{1 + \frac{2}{n}} + \ldots + \frac{1}{1 + \frac{n}{n}}$, which tends towards $\infty$ as $n \rightarrow \infty$

Since they both depend on $n$, if you try to treat their limits separately you will get $0 \times \infty$ which is an indeterminate form - i.e. you need to do some more work to find the actual limit.

It is actually a bit hard to evaluate this particular Riemann sum using normal limit methods (the easiest way to do it is to just prove that it is equal to the integral whose value you've already shown). However, if you choose a different subdivision of the interval $[0, 1]$ you can create a Riemann sum that you can evaluate to get the right answer, and with some changes of variable you can see the explanation in this question and its accepted answer.

Answer 4

Your error has nothing to do with anything specfic to integration, it's a common mistake of how to approch limits.

Let's make a more simple example.

$$\forall n>0: 1= \frac 1n \times n = \frac1n(1+\ldots +1);\;n \text{ summands } $$

Taking the limit for $n \to \infty$, using your approach we get

$$\lim_{n \to \infty} \frac1n(1+\ldots +1) = \frac1\infty(1+\ldots+1)=0,$$

which is obviously wrong.

Your first error is in writing $\frac1\infty$, which is not a meaningful expression. There is a reason your instructors likely have tried to prevent you from doing that, and the reason is that the relations and rules for calculations we "grow up with" no longer work when one of the operands is $\infty$. More on that in a moment.

A reasonable approach is to split the limit into a product of 2 limits:

$$\lim_{n \to \infty} \frac1n(1+\ldots +1) = \lim_{n \to \infty} \frac1n \times \lim_{n \to \infty}n.$$

That isn't totally correct, but "mostly". You can formally do that only if you know that the limits you split into exist. But since you are trying to solve a problem, that is a good approach. If it turns out they do exist, good! If it turns out one of them doesn't exist, then at least it was worth a try and we need to look for other measures.

So what about those limits?

Obviously, $\lim_{n \to \infty} \frac1n=0$, no problem.

Also, $\lim_{n \to \infty}n = \infty$, also obvisouly.

Now there is a reason that the sequence $a_n=n$ is called diverging. It doesn't converge to any real number, just like $-1,+1,-1,+1,\ldots$ isn't converging to any real number. Writing $\lim_{n \to \infty}n = \infty$ is a very nice shorthand, that can be justified with topological arguments, it gives you some information of the behavior of $a_n=n$ that is more specific than "diverging". But, to recap, the "limit" $\infty$ is not a number you can calculate with.

And now, with this example, we can see, why this is so. The above example shows that in this specific case, considering the limit we want to calculate (which is $1$) we "would like" to have $0 \times \infty = 1$.

But if we change the product, we may "want" different outcomes, for example for taking the following example to the limit

$$n=\frac1n n^2,$$

we again get that the first factor $\frac1n$ tends to $0$ when $n$ tends to $\infty$ and the second factor $n^2$ tends to $\infty$. So for this case, we want $0 \times \infty$ to be $\infty$.

And if you consider the limit of

$$\frac1n=\frac1{n^2}n,$$

we see that in that case we want $0 \times \infty$ to be $0$.

So, to recap: $0 \times \infty$ is not a meaningful real number, as we can see in simple cases, it "would be equal" to different values, depending on how exactly we arrived at those $0$ and $\infty$.

To come back to your original problem: Your product

$$ = \lim _{n \to \infty } (\frac{1}{n})( \frac {1} {1+\frac{1}{n}} + \frac {1} {1+\frac{2}{n}} + \frac {1} {1+\frac{3}{n}} + ... +\frac {1} {2} ) $$

is very similar to my initial example, your fractions in the sum are just a bit more complicated. What may have fooled you is that each term there is obviously bounded from above by $1$. The problem is that the $\ldots$ hide the fact that you have an increasing number of those summands, so any one being bounded means nothing.

Just look at my above initial of

$$\forall n>0: 1= \frac1n(1+\ldots +1);\;n \text{ summands, } $$

each summand is a constant, but you have $n$ summands, so their sum is $n$, and this not bounded.

Answer 5

You want to show that $$ \lim _{n \to \infty } \sum ^n _{i=1} \frac{1}{n+i}=\ln 2$$ As others have said, the usual way to do this is to define $$\ln x = \int_1^x \frac1{t} dt $$ and the workings in your question would then essentially complete the proof. If you want a proof that doesn't rely on the integral then we need a different definition for $\ln 2 $. Perhaps you are happy with the series: $$\ln 2=1-\frac12+\frac13-\frac14+ \cdots$$

in which case we can complete the proof without calculus as follows:

Note that $$\sum^n_{i \text{ odd}}\frac{1}{i} + \sum ^n_{i \text{ even}}\frac{1}{i}=\sum ^n_{i =1}\frac{1}{i}= 2\sum^{2n}_{i \text{ even}}\frac{1}{i} $$

We are interested in $$ \sum ^n _{i=1} \frac{1}{n+i}= \sum ^{2n}_{i=1} \frac{1}{i}-\sum ^{n}_{i=1} \frac{1}{i} $$ $$=\sum^{2n}_{i \text{ odd}}\frac{1}{i} + \sum ^{2n}_{i \text{ even}}\frac{1}{i} - 2 \sum^{2n}_{i \text{ even}}\frac{1}{i}$$ $$=\sum^{2n}_{i \text{ odd}}\frac{1}{i} - \sum ^{2n}_{i \text{ even}}\frac{1}{i} $$ but taking the limit we see this is equal to our definition of $\ln 2$.

Answer 6

Yes, it's possible to show both expressions equal $\ln(2)$. Just like how others pointed out, we can't just "plug in $\infty$" all the time and get the right answer because $\infty$ is not a real number. To show our Riemann Sum converges to $\ln{(2)}$, we do

$$\eqalign{ \lim_{n\to\infty}\sum_{i=1}^{n}\frac{1}{n+i} &= \lim_{n\to\infty}\sum_{i=1}^{n}\frac{1}{n\left(1+\frac{i}{n}\right)} \cr &= \lim_{n\to\infty}\sum_{i=1}^{n}\left(\frac{1}{1+\frac{i}{n}}\right)\left(\frac{1-0}{n}\right). }$$

Recall the definition of Riemann Sums on $\left[a,b\right]$ using $n$ subintervals that $x_i = a+i\Delta x$ and $\Delta x = \frac{b-a}{n}$. Let $b=1$ and $a=0$ so that $\Delta x = \frac{1-0}{n}$ and $x_i = 0 + i\left(\frac{1-0}{n}\right) = \frac{i}{n}$. Then we get the sum to equal

$$\int_0^1 \frac{1}{1+x}dx,$$

which equals $\ln{(2)}$.

Does that make sense?